Algebra Examples

$(1, 0)$ , $(4, 2)$ , $(3, - 7)$

Step 1

Use the standard form of a quadratic equation $y = a x^{2} + b x + c$ as the starting point for finding the equation through the three points.

$y = a x^{2} + b x + c$

Step 2

Create a system of equations by substituting the $x$ and $y$ values of each point into the standard formula of a quadratic equation to create the three equation system.

$0 = a {(1)}^{2} + b (1) + c, 2 = a {(4)}^{2} + b (4) + c, - 7 = a {(3)}^{2} + b (3) + c$

Step 3

Solve the system of equations to find the values of $a$ , $b$ , and $c$ .

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Step 3.1

Solve for $a$ in $0 = a + b + c$ .

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Step 3.1.1

Rewrite the equation as $a + b + c = 0$ .

$a + b + c = 0$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2

Move all terms not containing $a$ to the right side of the equation.

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Step 3.1.2.1

Subtract $b$ from both sides of the equation.

$a + c = - b$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2.2

Subtract $c$ from both sides of the equation.

$a = - b - c$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$a = - b - c$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$a = - b - c$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2

Replace all occurrences of $a$ with $- b - c$ in each equation.

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Step 3.2.1

Replace all occurrences of $a$ in $2 = a \cdot 4^{2} + b (4) + c$ with $- b - c$ .

$2 = (- b - c) \cdot 4^{2} + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $(- b - c) \cdot 4^{2} + b (4) + c$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Raise $4$ to the power of $2$ .

$2 = (- b - c) \cdot 16 + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.1.2

Apply the distributive property.

$2 = - b \cdot 16 - c \cdot 16 + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.1.3

Multiply $16$ by $- 1$ .

$2 = - 16 b - c \cdot 16 + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.1.4

Multiply $16$ by $- 1$ .

$2 = - 16 b - 16 c + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.1.5

Move $4$ to the left of $b$ .

$2 = - 16 b - 16 c + 4 b + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$2 = - 16 b - 16 c + 4 b + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.2

Simplify by adding terms.

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Step 3.2.2.1.2.1

Add $- 16 b$ and $4 b$ .

$2 = - 12 b - 16 c + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.2.2

Add $- 16 c$ and $c$ .

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.3

Replace all occurrences of $a$ in $- 7 = a \cdot 3^{2} + b (3) + c$ with $- b - c$ .

$- 7 = (- b - c) \cdot 3^{2} + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Simplify $(- b - c) \cdot 3^{2} + b (3) + c$ .

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Step 3.2.4.1.1

Simplify each term.

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Step 3.2.4.1.1.1

Raise $3$ to the power of $2$ .

$- 7 = (- b - c) \cdot 9 + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.1.2

Apply the distributive property.

$- 7 = - b \cdot 9 - c \cdot 9 + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.1.3

Multiply $9$ by $- 1$ .

$- 7 = - 9 b - c \cdot 9 + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.1.4

Multiply $9$ by $- 1$ .

$- 7 = - 9 b - 9 c + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.1.5

Move $3$ to the left of $b$ .

$- 7 = - 9 b - 9 c + 3 b + c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 9 b - 9 c + 3 b + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.2

Simplify by adding terms.

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Step 3.2.4.1.2.1

Add $- 9 b$ and $3 b$ .

$- 7 = - 6 b - 9 c + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.2.2

Add $- 9 c$ and $c$ .

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3

Solve for $b$ in $- 7 = - 6 b - 8 c$ .

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Step 3.3.1

Rewrite the equation as $- 6 b - 8 c = - 7$ .

$- 6 b - 8 c = - 7$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.2

Add $8 c$ to both sides of the equation.

$- 6 b = - 7 + 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3

Divide each term in $- 6 b = - 7 + 8 c$ by $- 6$ and simplify.

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Step 3.3.3.1

Divide each term in $- 6 b = - 7 + 8 c$ by $- 6$ .

$\frac{- 6 b}{- 6} = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $- 6$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Simplify each term.

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Step 3.3.3.3.1.1

Dividing two negative values results in a positive value.

$b = \frac{7}{6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.2

Cancel the common factor of $8$ and $- 6$ .

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Step 3.3.3.3.1.2.1

Factor $2$ out of $8 c$ .

$b = \frac{7}{6} + \frac{2 (4 c)}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.2.2

Cancel the common factors.

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Step 3.3.3.3.1.2.2.1

Factor $2$ out of $- 6$ .

$b = \frac{7}{6} + \frac{2 (4 c)}{2 (- 3)}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.2.2.2

Cancel the common factor.

$b = \frac{7}{6} + \frac{2 (4 c)}{2 \cdot - 3}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.2.2.3

Rewrite the expression.

$b = \frac{7}{6} + \frac{4 c}{- 3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} + \frac{4 c}{- 3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} + \frac{4 c}{- 3}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.3

Move the negative in front of the fraction.

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.4

Replace all occurrences of $b$ with $\frac{7}{6} - \frac{4 c}{3}$ in each equation.

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Step 3.4.1

Replace all occurrences of $b$ in $2 = - 12 b - 15 c$ with $\frac{7}{6} - \frac{4 c}{3}$ .

$2 = - 12 (\frac{7}{6} - \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 12 (\frac{7}{6} - \frac{4 c}{3}) - 15 c$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$2 = - 12 (\frac{7}{6}) - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.2

Cancel the common factor of $6$ .

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Step 3.4.2.1.1.2.1

Factor $6$ out of $- 12$ .

$2 = 6 (- 2) (\frac{7}{6}) - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.2.2

Cancel the common factor.

$2 = 6 \cdot (- 2 (\frac{7}{6})) - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.2.3

Rewrite the expression.

$2 = - 2 \cdot 7 - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 2 \cdot 7 - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.3

Multiply $- 2$ by $7$ .

$2 = - 14 - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.4

Cancel the common factor of $3$ .

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Step 3.4.2.1.1.4.1

Move the leading negative in $- \frac{4 c}{3}$ into the numerator.

$2 = - 14 - 12 \frac{- 4 c}{3} - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.4.2

Factor $3$ out of $- 12$ .

$2 = - 14 + 3 (- 4) (\frac{- 4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.4.3

Cancel the common factor.

$2 = - 14 + 3 \cdot (- 4 \frac{- 4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.4.4

Rewrite the expression.

$2 = - 14 - 4 (- 4 c) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 14 - 4 (- 4 c) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.5

Multiply $- 4$ by $- 4$ .

$2 = - 14 + 16 c - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 14 + 16 c - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.2

Subtract $15 c$ from $16 c$ .

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.3

Replace all occurrences of $b$ in $a = - b - c$ with $\frac{7}{6} - \frac{4 c}{3}$ .

$a = - (\frac{7}{6} - \frac{4 c}{3}) - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4

Simplify the right side.

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Step 3.4.4.1

Simplify $- (\frac{7}{6} - \frac{4 c}{3}) - c$ .

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Step 3.4.4.1.1

Simplify each term.

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Step 3.4.4.1.1.1

Apply the distributive property.

$a = - \frac{7}{6} + \frac{4 c}{3} - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.1.2

Multiply $- - \frac{4 c}{3}$ .

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Step 3.4.4.1.1.2.1

Multiply $- 1$ by $- 1$ .

$a = - \frac{7}{6} + 1 (\frac{4 c}{3}) - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.1.2.2

Multiply $\frac{4 c}{3}$ by $1$ .

$a = - \frac{7}{6} + \frac{4 c}{3} - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{4 c}{3} - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{4 c}{3} - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.2

To write $- c$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = - \frac{7}{6} + \frac{4 c}{3} - c \cdot \frac{3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.3

Simplify terms.

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Step 3.4.4.1.3.1

Combine $- c$ and $\frac{3}{3}$ .

$a = - \frac{7}{6} + \frac{4 c}{3} + \frac{- c \cdot 3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.3.2

Combine the numerators over the common denominator.

$a = - \frac{7}{6} + \frac{4 c - c \cdot 3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{4 c - c \cdot 3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4

Simplify each term.

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Step 3.4.4.1.4.1

Simplify the numerator.

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Step 3.4.4.1.4.1.1

Factor $c$ out of $4 c - c \cdot 3$ .

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Step 3.4.4.1.4.1.1.1

Factor $c$ out of $4 c$ .

$a = - \frac{7}{6} + \frac{c \cdot 4 - c \cdot 3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.1.1.2

Factor $c$ out of $- c \cdot 3$ .

$a = - \frac{7}{6} + \frac{c \cdot 4 + c (- 1 \cdot 3)}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.1.1.3

Factor $c$ out of $c \cdot 4 + c (- 1 \cdot 3)$ .

$a = - \frac{7}{6} + \frac{c (4 - 1 \cdot 3)}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c (4 - 1 \cdot 3)}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.1.2

Multiply $- 1$ by $3$ .

$a = - \frac{7}{6} + \frac{c (4 - 3)}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.1.3

Subtract $3$ from $4$ .

$a = - \frac{7}{6} + \frac{c \cdot 1}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c \cdot 1}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.2

Multiply $c$ by $1$ .

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.5

Solve for $c$ in $2 = - 14 + c$ .

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Step 3.5.1

Rewrite the equation as $- 14 + c = 2$ .

$- 14 + c = 2$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.5.2

Move all terms not containing $c$ to the right side of the equation.

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Step 3.5.2.1

Add $14$ to both sides of the equation.

$c = 2 + 14$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.5.2.2

Add $2$ and $14$ .

$c = 16$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

$c = 16$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

$c = 16$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6

Replace all occurrences of $c$ with $16$ in each equation.

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Step 3.6.1

Replace all occurrences of $c$ in $a = - \frac{7}{6} + \frac{c}{3}$ with $16$ .

$a = - \frac{7}{6} + \frac{16}{3}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $- \frac{7}{6} + \frac{16}{3}$ .

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Step 3.6.2.1.1

To write $\frac{16}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$a = - \frac{7}{6} + \frac{16}{3} \cdot \frac{2}{2}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.2

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.6.2.1.2.1

Multiply $\frac{16}{3}$ by $\frac{2}{2}$ .

$a = - \frac{7}{6} + \frac{16 \cdot 2}{3 \cdot 2}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.2.2

Multiply $3$ by $2$ .

$a = - \frac{7}{6} + \frac{16 \cdot 2}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{16 \cdot 2}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.3

Combine the numerators over the common denominator.

$a = \frac{- 7 + 16 \cdot 2}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.4

Simplify the numerator.

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Step 3.6.2.1.4.1

Multiply $16$ by $2$ .

$a = \frac{- 7 + 32}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.4.2

Add $- 7$ and $32$ .

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.3

Replace all occurrences of $c$ in $b = \frac{7}{6} - \frac{4 c}{3}$ with $16$ .

$b = \frac{7}{6} - \frac{4 (16)}{3}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Simplify $\frac{7}{6} - \frac{4 (16)}{3}$ .

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Step 3.6.4.1.1

Multiply $4$ by $16$ .

$b = \frac{7}{6} - \frac{64}{3}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.2

To write $- \frac{64}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$b = \frac{7}{6} - \frac{64}{3} \cdot \frac{2}{2}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.3

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.6.4.1.3.1

Multiply $\frac{64}{3}$ by $\frac{2}{2}$ .

$b = \frac{7}{6} - \frac{64 \cdot 2}{3 \cdot 2}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.3.2

Multiply $3$ by $2$ .

$b = \frac{7}{6} - \frac{64 \cdot 2}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{64 \cdot 2}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.4

Combine the numerators over the common denominator.

$b = \frac{7 - 64 \cdot 2}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.5

Simplify the numerator.

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Step 3.6.4.1.5.1

Multiply $- 64$ by $2$ .

$b = \frac{7 - 128}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.5.2

Subtract $128$ from $7$ .

$b = \frac{- 121}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{- 121}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.6

Move the negative in front of the fraction.

$b = - \frac{121}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = - \frac{121}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = - \frac{121}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = - \frac{121}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.7

List all of the solutions.

$b = - \frac{121}{6}, a = \frac{25}{6}, c = 16$

Step 4

Substitute the actual values of $a$ , $b$ , and $c$ into the formula for a quadratic equation to find the resulting equation.

$y = \frac{25 x^{2}}{6} - \frac{121 x}{6} + 16$

Step 5

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