Solve sqrt{6c+15}=c+4

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

Solve for c

$$c=-1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{6c+15} to the power of 2 and get 6c+15.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(c+4\right)^{2}.

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

Subtract c^{2} from both sides.

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

Subtract 8c from both sides.

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

Combine 6c and -8c to get -2c.

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

Subtract 16 from both sides.

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

Subtract 16 from 15 to get -1.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -c^{2}+ac+bc-1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-\left(-1\right)=1$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-1$$

Rewrite -c^{2}-2c-1 as \left(-c^{2}-c\right)+\left(-c-1\right).

$$\left(-c^{2}-c\right)+\left(-c-1\right)$$

Factor out c in -c^{2}-c.

$$c\left(-c-1\right)-c-1$$

Factor out common term -c-1 by using distributive property.

$$\left(-c-1\right)\left(c+1\right)$$

To find equation solutions, solve -c-1=0 and c+1=0.

$$c=-1 c=-1$$

Substitute -1 for c in the equation \sqrt{6c+15}=c+4.

$$\sqrt{6\left(-1\right)+15}=-1+4$$

Simplify. The value c=-1 satisfies the equation.

$$3=3$$

Substitute -1 for c in the equation \sqrt{6c+15}=c+4.

$$\sqrt{6\left(-1\right)+15}=-1+4$$

Simplify. The value c=-1 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{6c+15}=c+4.

$$c=-1 c=-1$$