Solve for c square root of 6c+15=c+4

$\sqrt{6 c + 15} = c + 4$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{6 c + 15}}^{2} = {(c + 4)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{6 c + 15}$ as ${(6 c + 15)}^{\frac{1}{2}}$ .

${({(6 c + 15)}^{\frac{1}{2}})}^{2} = {(c + 4)}^{2}$

Simplify the left side.

Simplify ${({(6 c + 15)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(6 c + 15)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(6 c + 15)}^{\frac{1}{2} \cdot 2} = {(c + 4)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(6 c + 15)}^{\frac{1}{2} \cdot 2} = {(c + 4)}^{2}$

Rewrite the expression.

${(6 c + 15)}^{1} = {(c + 4)}^{2}$

Simplify.

$6 c + 15 = {(c + 4)}^{2}$

Simplify the right side.

Simplify ${(c + 4)}^{2}$ .

Rewrite ${(c + 4)}^{2}$ as $(c + 4) (c + 4)$ .

$6 c + 15 = (c + 4) (c + 4)$

Expand $(c + 4) (c + 4)$ using the FOIL Method.

Apply the distributive property.

$6 c + 15 = c (c + 4) + 4 (c + 4)$

Apply the distributive property.

$6 c + 15 = c \cdot c + c \cdot 4 + 4 (c + 4)$

Apply the distributive property.

$6 c + 15 = c \cdot c + c \cdot 4 + 4 c + 4 \cdot 4$

Simplify and combine like terms.

Simplify each term.

Multiply $c$ by $c$ .

$6 c + 15 = c^{2} + c \cdot 4 + 4 c + 4 \cdot 4$

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

$6 c + 15 = c^{2} + 4 \cdot c + 4 c + 4 \cdot 4$

Multiply $4$ by $4$ .

$6 c + 15 = c^{2} + 4 c + 4 c + 16$

Add $4 c$ and $4 c$ .

$6 c + 15 = c^{2} + 8 c + 16$

Solve for $c$ .

Since $c$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$c^{2} + 8 c + 16 = 6 c + 15$

Move all terms containing $c$ to the left side of the equation.

Subtract $6 c$ from both sides of the equation.

$c^{2} + 8 c + 16 - 6 c = 15$

Subtract $6 c$ from $8 c$ .

$c^{2} + 2 c + 16 = 15$

Subtract $15$ from both sides of the equation.

$c^{2} + 2 c + 16 - 15 = 0$

Subtract $15$ from $16$ .

$c^{2} + 2 c + 1 = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$c^{2} + 2 c + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 c = 2 \cdot c \cdot 1$

Rewrite the polynomial.

$c^{2} + 2 \cdot c \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = c$ and $b = 1$ .

${(c + 1)}^{2} = 0$

Set the $c + 1$ equal to $0$ .

$c + 1 = 0$

Subtract $1$ from both sides of the equation.

$c = - 1$

Solve for c square root of 6c+15=c+4

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