Solve x^2+6x+c=(x+h)^2

$$x ^ { 2 } + 6 x + c = ( x + h ) ^ { 2 }$$

Solve for c

$$c=2hx-6x+h^{2}$$

Solution Steps

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

$$x^{2}+6x+c=x^{2}+2xh+h^{2}$$

Subtract x^{2} from both sides.

$$6x+c=x^{2}+2xh+h^{2}-x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$6x+c=2xh+h^{2}$$

Subtract 6x from both sides.

$$c=2xh+h^{2}-6x$$

Solve for h (complex solution)

$$h=-\left(\sqrt{x^{2}+6x+c}+x\right)$$

$$h=\sqrt{x^{2}+6x+c}-x$$

Solve for h

$$h=-\left(\sqrt{x^{2}+6x+c}+x\right)$$

$$h=\sqrt{x^{2}+6x+c}-x\text{, }c\geq -\left(x^{2}+6x\right)$$