Solve a+b^2=(a-b)^2

$$a + b ^ { 2 } = ( a - b ) ^ { 2 }$$

Solve for b

$$\left\{\begin{matrix}\\b=\frac{a-1}{2}\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$a+b^{2}=a^{2}-2ab+b^{2}$$

Add 2ab to both sides.

$$a+b^{2}+2ab=a^{2}+b^{2}$$

Subtract b^{2} from both sides.

$$a+b^{2}+2ab-b^{2}=a^{2}$$

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

$$a+2ab=a^{2}$$

Subtract a from both sides.

$$2ab=a^{2}-a$$

Divide both sides by 2a.

$$\frac{2ab}{2a}=\frac{a\left(a-1\right)}{2a}$$

Dividing by 2a undoes the multiplication by 2a.

$$b=\frac{a\left(a-1\right)}{2a}$$

Divide a\left(-1+a\right) by 2a.

$$b=\frac{a-1}{2}$$

Solve for a

$$a=2b+1$$

$$a=0$$