Finite Math Examples

${(\frac{2 c - 3}{5})}^{2} + 2 (\frac{2 c - 3}{5}) = 8$

Step 1

Simplify ${(\frac{2 c - 3}{5})}^{2} + 2 (\frac{2 c - 3}{5})$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the product rule to $\frac{2 c - 3}{5}$ .

$\frac{{(2 c - 3)}^{2}}{5^{2}} + 2 (\frac{2 c - 3}{5}) = 8$

Step 1.1.2

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

$\frac{{(2 c - 3)}^{2}}{25} + 2 (\frac{2 c - 3}{5}) = 8$

Step 1.1.3

Combine $2$ and $\frac{2 c - 3}{5}$ .

$\frac{{(2 c - 3)}^{2}}{25} + \frac{2 (2 c - 3)}{5} = 8$

Step 1.2

To write $\frac{2 (2 c - 3)}{5}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{{(2 c - 3)}^{2}}{25} + \frac{2 (2 c - 3)}{5} \cdot \frac{5}{5} = 8$

Step 1.3

Write each expression with a common denominator of $25$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{2 (2 c - 3)}{5}$ by $\frac{5}{5}$ .

$\frac{{(2 c - 3)}^{2}}{25} + \frac{2 (2 c - 3) \cdot 5}{5 \cdot 5} = 8$

Step 1.3.2

Multiply $5$ by $5$ .

$\frac{{(2 c - 3)}^{2}}{25} + \frac{2 (2 c - 3) \cdot 5}{25} = 8$

Step 1.4

Combine the numerators over the common denominator.

$\frac{{(2 c - 3)}^{2} + 2 (2 c - 3) \cdot 5}{25} = 8$

Step 1.5

Simplify the numerator.

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Step 1.5.1

Factor $2 c - 3$ out of ${(2 c - 3)}^{2} + 2 (2 c - 3) \cdot 5$ .

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Step 1.5.1.1

Factor $2 c - 3$ out of ${(2 c - 3)}^{2}$ .

$\frac{(2 c - 3) (2 c - 3) + 2 (2 c - 3) \cdot 5}{25} = 8$

Step 1.5.1.2

Factor $2 c - 3$ out of $2 (2 c - 3) \cdot 5$ .

$\frac{(2 c - 3) (2 c - 3) + (2 c - 3) (2 \cdot 5)}{25} = 8$

Step 1.5.1.3

Factor $2 c - 3$ out of $(2 c - 3) (2 c - 3) + (2 c - 3) (2 \cdot 5)$ .

$\frac{(2 c - 3) (2 c - 3 + 2 \cdot 5)}{25} = 8$

Step 1.5.2

Multiply $2$ by $5$ .

$\frac{(2 c - 3) (2 c - 3 + 10)}{25} = 8$

Step 1.5.3

Add $- 3$ and $10$ .

$\frac{(2 c - 3) (2 c + 7)}{25} = 8$

Step 2

Multiply both sides by $25$ .

$\frac{(2 c - 3) (2 c + 7)}{25} \cdot 25 = 8 \cdot 25$

Step 3

Simplify.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify $\frac{(2 c - 3) (2 c + 7)}{25} \cdot 25$ .

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Step 3.1.1.1

Cancel the common factor of $25$ .

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Step 3.1.1.1.1

Cancel the common factor.

$\frac{(2 c - 3) (2 c + 7)}{25} \cdot 25 = 8 \cdot 25$

Step 3.1.1.1.2

Rewrite the expression.

$(2 c - 3) (2 c + 7) = 8 \cdot 25$

Step 3.1.1.2

Expand $(2 c - 3) (2 c + 7)$ using the FOIL Method.

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Step 3.1.1.2.1

Apply the distributive property.

$2 c (2 c + 7) - 3 (2 c + 7) = 8 \cdot 25$

Step 3.1.1.2.2

Apply the distributive property.

$2 c (2 c) + 2 c \cdot 7 - 3 (2 c + 7) = 8 \cdot 25$

Step 3.1.1.2.3

Apply the distributive property.

$2 c (2 c) + 2 c \cdot 7 - 3 (2 c) - 3 \cdot 7 = 8 \cdot 25$

Step 3.1.1.3

Simplify and combine like terms.

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Step 3.1.1.3.1

Simplify each term.

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Step 3.1.1.3.1.1

Rewrite using the commutative property of multiplication.

$2 \cdot 2 c \cdot c + 2 c \cdot 7 - 3 (2 c) - 3 \cdot 7 = 8 \cdot 25$

Step 3.1.1.3.1.2

Multiply $c$ by $c$ by adding the exponents.

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Step 3.1.1.3.1.2.1

Move $c$ .

$2 \cdot 2 (c \cdot c) + 2 c \cdot 7 - 3 (2 c) - 3 \cdot 7 = 8 \cdot 25$

Step 3.1.1.3.1.2.2

Multiply $c$ by $c$ .

$2 \cdot 2 c^{2} + 2 c \cdot 7 - 3 (2 c) - 3 \cdot 7 = 8 \cdot 25$

Step 3.1.1.3.1.3

Multiply $2$ by $2$ .

$4 c^{2} + 2 c \cdot 7 - 3 (2 c) - 3 \cdot 7 = 8 \cdot 25$

Step 3.1.1.3.1.4

Multiply $7$ by $2$ .

$4 c^{2} + 14 c - 3 (2 c) - 3 \cdot 7 = 8 \cdot 25$

Step 3.1.1.3.1.5

Multiply $2$ by $- 3$ .

$4 c^{2} + 14 c - 6 c - 3 \cdot 7 = 8 \cdot 25$

Step 3.1.1.3.1.6

Multiply $- 3$ by $7$ .

$4 c^{2} + 14 c - 6 c - 21 = 8 \cdot 25$

Step 3.1.1.3.2

Subtract $6 c$ from $14 c$ .

$4 c^{2} + 8 c - 21 = 8 \cdot 25$

Step 3.2

Simplify the right side.

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Step 3.2.1

Multiply $8$ by $25$ .

$4 c^{2} + 8 c - 21 = 200$

Step 4

Solve for $c$ .

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Step 4.1

Subtract $200$ from both sides of the equation.

$4 c^{2} + 8 c - 21 - 200 = 0$

Step 4.2

Subtract $200$ from $- 21$ .

$4 c^{2} + 8 c - 221 = 0$

Step 4.3

Factor by grouping.

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Step 4.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot - 221 = - 884$ and whose sum is $b = 8$ .

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Step 4.3.1.1

Factor $8$ out of $8 c$ .

$4 c^{2} + 8 (c) - 221 = 0$

Step 4.3.1.2

Rewrite $8$ as $- 26$ plus $34$

$4 c^{2} + (- 26 + 34) c - 221 = 0$

Step 4.3.1.3

Apply the distributive property.

$4 c^{2} - 26 c + 34 c - 221 = 0$

Step 4.3.2

Factor out the greatest common factor from each group.

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Step 4.3.2.1

Group the first two terms and the last two terms.

$(4 c^{2} - 26 c) + 34 c - 221 = 0$

Step 4.3.2.2

Factor out the greatest common factor (GCF) from each group.

$2 c (2 c - 13) + 17 (2 c - 13) = 0$

Step 4.3.3

Factor the polynomial by factoring out the greatest common factor, $2 c - 13$ .

$(2 c - 13) (2 c + 17) = 0$

Step 4.4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 c - 13 = 0$

$2 c + 17 = 0$

Step 4.5

Set $2 c - 13$ equal to $0$ and solve for $c$ .

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Step 4.5.1

Set $2 c - 13$ equal to $0$ .

$2 c - 13 = 0$

Step 4.5.2

Solve $2 c - 13 = 0$ for $c$ .

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Step 4.5.2.1

Add $13$ to both sides of the equation.

$2 c = 13$

Step 4.5.2.2

Divide each term in $2 c = 13$ by $2$ and simplify.

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Step 4.5.2.2.1

Divide each term in $2 c = 13$ by $2$ .

$\frac{2 c}{2} = \frac{13}{2}$

Step 4.5.2.2.2

Simplify the left side.

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Step 4.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 c}{2} = \frac{13}{2}$

Step 4.5.2.2.2.1.2

Divide $c$ by $1$ .

$c = \frac{13}{2}$

Step 4.6

Set $2 c + 17$ equal to $0$ and solve for $c$ .

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Step 4.6.1

Set $2 c + 17$ equal to $0$ .

$2 c + 17 = 0$

Step 4.6.2

Solve $2 c + 17 = 0$ for $c$ .

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Step 4.6.2.1

Subtract $17$ from both sides of the equation.

$2 c = - 17$

Step 4.6.2.2

Divide each term in $2 c = - 17$ by $2$ and simplify.

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Step 4.6.2.2.1

Divide each term in $2 c = - 17$ by $2$ .

$\frac{2 c}{2} = \frac{- 17}{2}$

Step 4.6.2.2.2

Simplify the left side.

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Step 4.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.6.2.2.2.1.1

Cancel the common factor.

$\frac{2 c}{2} = \frac{- 17}{2}$

Step 4.6.2.2.2.1.2

Divide $c$ by $1$ .

$c = \frac{- 17}{2}$

Step 4.6.2.2.3

Simplify the right side.

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Step 4.6.2.2.3.1

Move the negative in front of the fraction.

$c = - \frac{17}{2}$

Step 4.7

The final solution is all the values that make $(2 c - 13) (2 c + 17) = 0$ true.

$c = \frac{13}{2}, - \frac{17}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$c = \frac{13}{2}, - \frac{17}{2}$

Decimal Form:

$c = 6.5, - 8.5$

Mixed Number Form:

$c = 6 \frac{1}{2}, - 8 \frac{1}{2}$