Solve -left(x-1right)^2+x^2+2x+1+2x=x^2+hx+1

$$- { \left(x-1 \right) }^{ 2 } + { x }^{ 2 } +2x+1+2x = { x }^{ 2 } +hx+1$$

Solve for h

$$h=-x+6-\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$-\left(x^{2}-2x+1\right)+x^{2}+2x+1+2x=x^{2}+hx+1$$

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$-x^{2}+2x-1+x^{2}+2x+1+2x=x^{2}+hx+1$$

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

$$2x-1+2x+1+2x=x^{2}+hx+1$$

Combine 2x and 2x to get 4x.

$$4x-1+1+2x=x^{2}+hx+1$$

Add -1 and 1 to get 0.

$$4x+2x=x^{2}+hx+1$$

Combine 4x and 2x to get 6x.

$$6x=x^{2}+hx+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+hx+1=6x$$

Subtract x^{2} from both sides.

$$hx+1=6x-x^{2}$$

Subtract 1 from both sides.

$$hx=6x-x^{2}-1$$

The equation is in standard form.

$$xh=-x^{2}+6x-1$$

Divide both sides by x.

$$\frac{xh}{x}=\frac{-x^{2}+6x-1}{x}$$

Dividing by x undoes the multiplication by x.

$$h=\frac{-x^{2}+6x-1}{x}$$

Divide 6x-x^{2}-1 by x.

$$h=-x+6-\frac{1}{x}$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{\left(h-8\right)\left(h-4\right)}-h+6}{2}$$

$$x=\frac{\sqrt{\left(h-8\right)\left(h-4\right)}-h+6}{2}$$

Solve for x

$$x=\frac{-\sqrt{\left(h-8\right)\left(h-4\right)}-h+6}{2}$$

$$x=\frac{\sqrt{\left(h-8\right)\left(h-4\right)}-h+6}{2}\text{, }h\leq 4\text{ or }h\geq 8$$