Solve (x-h)(x+h)+(h+1)^2=(x-h)^2+2(x+1)

$$( x - h ) ( x + h ) + ( h + 1 ) ^ { 2 } = ( x - h ) ^ { 2 } + 2 ( x + 1 )$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{h-1}{2}\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&h=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Consider \left(x-h\right)\left(x+h\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

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

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

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

Use the distributive property to multiply 2 by x+1.

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

Subtract x^{2} from both sides.

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

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

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

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

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

Subtract h^{2} from both sides.

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

Subtract 2 from both sides.

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

Subtract 2 from 1 to get -1.

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

Combine all terms containing x.

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

The equation is in standard form.

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

Divide both sides by -2h+2.

$$\frac{\left(2-2h\right)x}{2-2h}=-\frac{\left(h-1\right)^{2}}{2-2h}$$

Dividing by -2h+2 undoes the multiplication by -2h+2.

$$x=-\frac{\left(h-1\right)^{2}}{2-2h}$$

Divide -\left(h-1\right)^{2} by -2h+2.

$$x=\frac{h-1}{2}$$

Solve for x

$$\left\{\begin{matrix}\\x=\frac{h-1}{2}\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&h=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Consider \left(x-h\right)\left(x+h\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

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

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

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

Use the distributive property to multiply 2 by x+1.

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

Subtract x^{2} from both sides.

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

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

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

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

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

Subtract h^{2} from both sides.

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

Subtract 2 from both sides.

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

Subtract 2 from 1 to get -1.

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

Combine all terms containing x.

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

The equation is in standard form.

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

Divide both sides by -2h+2.

$$\frac{\left(2-2h\right)x}{2-2h}=-\frac{\left(h-1\right)^{2}}{2-2h}$$

Dividing by -2h+2 undoes the multiplication by -2h+2.

$$x=-\frac{\left(h-1\right)^{2}}{2-2h}$$

Divide -\left(h-1\right)^{2} by -2h+2.

$$x=\frac{h-1}{2}$$

Solve for h

$$h=1$$

$$h=2x+1$$