Precalculus Examples

$(5, 3)$ , $(6, - 9)$ , $(- 4, 1)$

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.

$3 = a {(5)}^{2} + b (5) + c, - 9 = a {(6)}^{2} + b (6) + c, 1 = a {(- 4)}^{2} + b (- 4) + c$

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 $c$ in $3 = a \cdot 5^{2} + b (5) + c$ .

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Step 3.1.1

Rewrite the equation as $a \cdot 5^{2} + b (5) + c = 3$ .

$a \cdot 5^{2} + b (5) + c = 3$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.1.2

Simplify each term.

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Step 3.1.2.1

Raise $5$ to the power of $2$ .

$a \cdot 25 + b (5) + c = 3$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.1.2.2

Move $25$ to the left of $a$ .

$25 \cdot a + b (5) + c = 3$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.1.2.3

Move $5$ to the left of $b$ .

$25 a + 5 b + c = 3$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$25 a + 5 b + c = 3$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.1.3

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

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Step 3.1.3.1

Subtract $25 a$ from both sides of the equation.

$5 b + c = 3 - 25 a$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.1.3.2

Subtract $5 b$ from both sides of the equation.

$c = 3 - 25 a - 5 b$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$c = 3 - 25 a - 5 b$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$c = 3 - 25 a - 5 b$

$- 9 = a \cdot 6^{2} + b (6) + c$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.2

Replace all occurrences of $c$ with $3 - 25 a - 5 b$ in each equation.

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Step 3.2.1

Replace all occurrences of $c$ in $- 9 = a \cdot 6^{2} + b (6) + c$ with $3 - 25 a - 5 b$ .

$- 9 = a \cdot 6^{2} + b (6) + 3 - 25 a - 5 b$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.2.2

Simplify $- 9 = a \cdot 6^{2} + b (6) + 3 - 25 a - 5 b$ .

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Remove parentheses.

$- 9 = a \cdot 6^{2} + b (6) + 3 - 25 a - 5 b$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$- 9 = a \cdot 6^{2} + b (6) + 3 - 25 a - 5 b$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Simplify $a \cdot 6^{2} + b (6) + 3 - 25 a - 5 b$ .

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Step 3.2.2.2.1.1

Simplify each term.

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Step 3.2.2.2.1.1.1

Raise $6$ to the power of $2$ .

$- 9 = a \cdot 36 + b (6) + 3 - 25 a - 5 b$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.2.2.2.1.1.2

Move $36$ to the left of $a$ .

$- 9 = 36 \cdot a + b (6) + 3 - 25 a - 5 b$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.2.2.2.1.1.3

Move $6$ to the left of $b$ .

$- 9 = 36 a + 6 b + 3 - 25 a - 5 b$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$- 9 = 36 a + 6 b + 3 - 25 a - 5 b$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.2.2.2.1.2

Simplify by adding terms.

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Step 3.2.2.2.1.2.1

Subtract $25 a$ from $36 a$ .

$- 9 = 11 a + 6 b + 3 - 5 b$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.2.2.2.1.2.2

Subtract $5 b$ from $6 b$ .

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + c$

Step 3.2.3

Replace all occurrences of $c$ in $1 = a {(- 4)}^{2} + b (- 4) + c$ with $3 - 25 a - 5 b$ .

$1 = a {(- 4)}^{2} + b (- 4) + 3 - 25 a - 5 b$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

Step 3.2.4

Simplify $1 = a {(- 4)}^{2} + b (- 4) + 3 - 25 a - 5 b$ .

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Step 3.2.4.1

Simplify the left side.

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Step 3.2.4.1.1

Remove parentheses.

$1 = a {(- 4)}^{2} + b (- 4) + 3 - 25 a - 5 b$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = a {(- 4)}^{2} + b (- 4) + 3 - 25 a - 5 b$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

Step 3.2.4.2

Simplify the right side.

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Step 3.2.4.2.1

Simplify $a {(- 4)}^{2} + b (- 4) + 3 - 25 a - 5 b$ .

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Step 3.2.4.2.1.1

Simplify each term.

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Step 3.2.4.2.1.1.1

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

$1 = a \cdot 16 + b (- 4) + 3 - 25 a - 5 b$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

Step 3.2.4.2.1.1.2

Move $16$ to the left of $a$ .

$1 = 16 \cdot a + b (- 4) + 3 - 25 a - 5 b$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

Step 3.2.4.2.1.1.3

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

$1 = 16 a - 4 b + 3 - 25 a - 5 b$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = 16 a - 4 b + 3 - 25 a - 5 b$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

Step 3.2.4.2.1.2

Simplify by adding terms.

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Step 3.2.4.2.1.2.1

Subtract $25 a$ from $16 a$ .

$1 = - 9 a - 4 b + 3 - 5 b$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

Step 3.2.4.2.1.2.2

Subtract $5 b$ from $- 4 b$ .

$1 = - 9 a - 9 b + 3$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = - 9 a - 9 b + 3$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = - 9 a - 9 b + 3$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = - 9 a - 9 b + 3$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = - 9 a - 9 b + 3$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

$1 = - 9 a - 9 b + 3$

$- 9 = 11 a + b + 3$

$c = 3 - 25 a - 5 b$

Step 3.3

Solve for $b$ in $- 9 = 11 a + b + 3$ .

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Step 3.3.1

Rewrite the equation as $11 a + b + 3 = - 9$ .

$11 a + b + 3 = - 9$

$1 = - 9 a - 9 b + 3$

$c = 3 - 25 a - 5 b$

Step 3.3.2

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

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Step 3.3.2.1

Subtract $11 a$ from both sides of the equation.

$b + 3 = - 9 - 11 a$

$1 = - 9 a - 9 b + 3$

$c = 3 - 25 a - 5 b$

Step 3.3.2.2

Subtract $3$ from both sides of the equation.

$b = - 9 - 11 a - 3$

$1 = - 9 a - 9 b + 3$

$c = 3 - 25 a - 5 b$

Step 3.3.2.3

Subtract $3$ from $- 9$ .

$b = - 11 a - 12$

$1 = - 9 a - 9 b + 3$

$c = 3 - 25 a - 5 b$

$b = - 11 a - 12$

$1 = - 9 a - 9 b + 3$

$c = 3 - 25 a - 5 b$

$b = - 11 a - 12$

$1 = - 9 a - 9 b + 3$

$c = 3 - 25 a - 5 b$

Step 3.4

Replace all occurrences of $b$ with $- 11 a - 12$ in each equation.

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Step 3.4.1

Replace all occurrences of $b$ in $1 = - 9 a - 9 b + 3$ with $- 11 a - 12$ .

$1 = - 9 a - 9 (- 11 a - 12) + 3$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 9 a - 9 (- 11 a - 12) + 3$ .

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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.

$1 = - 9 a - 9 (- 11 a) - 9 \cdot - 12 + 3$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

Step 3.4.2.1.1.2

Multiply $- 11$ by $- 9$ .

$1 = - 9 a + 99 a - 9 \cdot - 12 + 3$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

Step 3.4.2.1.1.3

Multiply $- 9$ by $- 12$ .

$1 = - 9 a + 99 a + 108 + 3$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

$1 = - 9 a + 99 a + 108 + 3$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

Step 3.4.2.1.2

Simplify by adding terms.

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Step 3.4.2.1.2.1

Add $- 9 a$ and $99 a$ .

$1 = 90 a + 108 + 3$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

Step 3.4.2.1.2.2

Add $108$ and $3$ .

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 3 - 25 a - 5 b$

Step 3.4.3

Replace all occurrences of $b$ in $c = 3 - 25 a - 5 b$ with $- 11 a - 12$ .

$c = 3 - 25 a - 5 (- 11 a - 12)$

$1 = 90 a + 111$

$b = - 11 a - 12$

Step 3.4.4

Simplify the right side.

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Step 3.4.4.1

Simplify $3 - 25 a - 5 (- 11 a - 12)$ .

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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.

$c = 3 - 25 a - 5 (- 11 a) - 5 \cdot - 12$

$1 = 90 a + 111$

$b = - 11 a - 12$

Step 3.4.4.1.1.2

Multiply $- 11$ by $- 5$ .

$c = 3 - 25 a + 55 a - 5 \cdot - 12$

$1 = 90 a + 111$

$b = - 11 a - 12$

Step 3.4.4.1.1.3

Multiply $- 5$ by $- 12$ .

$c = 3 - 25 a + 55 a + 60$

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 3 - 25 a + 55 a + 60$

$1 = 90 a + 111$

$b = - 11 a - 12$

Step 3.4.4.1.2

Simplify by adding terms.

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Step 3.4.4.1.2.1

Add $3$ and $60$ .

$c = - 25 a + 55 a + 63$

$1 = 90 a + 111$

$b = - 11 a - 12$

Step 3.4.4.1.2.2

Add $- 25 a$ and $55 a$ .

$c = 30 a + 63$

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 30 a + 63$

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 30 a + 63$

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 30 a + 63$

$1 = 90 a + 111$

$b = - 11 a - 12$

$c = 30 a + 63$

$1 = 90 a + 111$

$b = - 11 a - 12$

Step 3.5

Solve for $a$ in $1 = 90 a + 111$ .

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Step 3.5.1

Rewrite the equation as $90 a + 111 = 1$ .

$90 a + 111 = 1$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.2

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

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Step 3.5.2.1

Subtract $111$ from both sides of the equation.

$90 a = 1 - 111$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.2.2

Subtract $111$ from $1$ .

$90 a = - 110$

$c = 30 a + 63$

$b = - 11 a - 12$

$90 a = - 110$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.3

Divide each term in $90 a = - 110$ by $90$ and simplify.

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Step 3.5.3.1

Divide each term in $90 a = - 110$ by $90$ .

$\frac{90 a}{90} = \frac{- 110}{90}$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $90$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{90 a}{90} = \frac{- 110}{90}$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 110}{90}$

$c = 30 a + 63$

$b = - 11 a - 12$

$a = \frac{- 110}{90}$

$c = 30 a + 63$

$b = - 11 a - 12$

$a = \frac{- 110}{90}$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Cancel the common factor of $- 110$ and $90$ .

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Step 3.5.3.3.1.1

Factor $10$ out of $- 110$ .

$a = \frac{10 (- 11)}{90}$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.3.3.1.2

Cancel the common factors.

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Step 3.5.3.3.1.2.1

Factor $10$ out of $90$ .

$a = \frac{10 \cdot - 11}{10 \cdot 9}$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.3.3.1.2.2

Cancel the common factor.

$a = \frac{10 \cdot - 11}{10 \cdot 9}$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.3.3.1.2.3

Rewrite the expression.

$a = \frac{- 11}{9}$

$c = 30 a + 63$

$b = - 11 a - 12$

$a = \frac{- 11}{9}$

$c = 30 a + 63$

$b = - 11 a - 12$

$a = \frac{- 11}{9}$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.5.3.3.2

Move the negative in front of the fraction.

$a = - \frac{11}{9}$

$c = 30 a + 63$

$b = - 11 a - 12$

$a = - \frac{11}{9}$

$c = 30 a + 63$

$b = - 11 a - 12$

$a = - \frac{11}{9}$

$c = 30 a + 63$

$b = - 11 a - 12$

$a = - \frac{11}{9}$

$c = 30 a + 63$

$b = - 11 a - 12$

Step 3.6

Replace all occurrences of $a$ with $- \frac{11}{9}$ in each equation.

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Step 3.6.1

Replace all occurrences of $a$ in $c = 30 a + 63$ with $- \frac{11}{9}$ .

$c = 30 (- \frac{11}{9}) + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $30 (- \frac{11}{9}) + 63$ .

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Step 3.6.2.1.1

Simplify each term.

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Step 3.6.2.1.1.1

Cancel the common factor of $3$ .

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Step 3.6.2.1.1.1.1

Move the leading negative in $- \frac{11}{9}$ into the numerator.

$c = 30 (\frac{- 11}{9}) + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.1.1.2

Factor $3$ out of $30$ .

$c = 3 (10) (\frac{- 11}{9}) + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.1.1.3

Factor $3$ out of $9$ .

$c = 3 \cdot (10 (\frac{- 11}{3 \cdot 3})) + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.1.1.4

Cancel the common factor.

$c = 3 \cdot (10 (\frac{- 11}{3 \cdot 3})) + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.1.1.5

Rewrite the expression.

$c = 10 (\frac{- 11}{3}) + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

$c = 10 (\frac{- 11}{3}) + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.1.2

Combine $10$ and $\frac{- 11}{3}$ .

$c = \frac{10 \cdot - 11}{3} + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.1.3

Multiply $10$ by $- 11$ .

$c = \frac{- 110}{3} + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.1.4

Move the negative in front of the fraction.

$c = - \frac{110}{3} + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

$c = - \frac{110}{3} + 63$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.2

To write $63$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$c = - \frac{110}{3} + 63 \cdot \frac{3}{3}$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.3

Combine $63$ and $\frac{3}{3}$ .

$c = - \frac{110}{3} + \frac{63 \cdot 3}{3}$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.4

Combine the numerators over the common denominator.

$c = \frac{- 110 + 63 \cdot 3}{3}$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.5

Simplify the numerator.

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Step 3.6.2.1.5.1

Multiply $63$ by $3$ .

$c = \frac{- 110 + 189}{3}$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.2.1.5.2

Add $- 110$ and $189$ .

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - 11 a - 12$

Step 3.6.3

Replace all occurrences of $a$ in $b = - 11 a - 12$ with $- \frac{11}{9}$ .

$b = - 11 (- \frac{11}{9}) - 12$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Simplify $- 11 (- \frac{11}{9}) - 12$ .

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Step 3.6.4.1.1

Multiply $- 11 (- \frac{11}{9})$ .

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Step 3.6.4.1.1.1

Multiply $- 1$ by $- 11$ .

$b = 11 (\frac{11}{9}) - 12$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.6.4.1.1.2

Combine $11$ and $\frac{11}{9}$ .

$b = \frac{11 \cdot 11}{9} - 12$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.6.4.1.1.3

Multiply $11$ by $11$ .

$b = \frac{121}{9} - 12$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{121}{9} - 12$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.6.4.1.2

To write $- 12$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$b = \frac{121}{9} - 12 \cdot \frac{9}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.6.4.1.3

Combine $- 12$ and $\frac{9}{9}$ .

$b = \frac{121}{9} + \frac{- 12 \cdot 9}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.6.4.1.4

Combine the numerators over the common denominator.

$b = \frac{121 - 12 \cdot 9}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.6.4.1.5

Simplify the numerator.

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Step 3.6.4.1.5.1

Multiply $- 12$ by $9$ .

$b = \frac{121 - 108}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.6.4.1.5.2

Subtract $108$ from $121$ .

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

Step 3.7

List all of the solutions.

$b = \frac{13}{9}, c = \frac{79}{3}, a = - \frac{11}{9}$

Step 4

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

$y = - \frac{11 x^{2}}{9} + \frac{13 x}{9} + \frac{79}{3}$

Step 5

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