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

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

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{a+1}{4}\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$4x^{2}+4xa+a^{2}-\left(2x-a\right)^{2}-2a=2a^{2}$$

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

$$4x^{2}+4xa+a^{2}-\left(4x^{2}-4xa+a^{2}\right)-2a=2a^{2}$$

To find the opposite of 4x^{2}-4xa+a^{2}, find the opposite of each term.

$$4x^{2}+4xa+a^{2}-4x^{2}+4xa-a^{2}-2a=2a^{2}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$4xa+a^{2}+4xa-a^{2}-2a=2a^{2}$$

Combine 4xa and 4xa to get 8xa.

$$8xa+a^{2}-a^{2}-2a=2a^{2}$$

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

$$8xa-2a=2a^{2}$$

Add 2a to both sides.

$$8xa=2a^{2}+2a$$

The equation is in standard form.

$$8ax=2a^{2}+2a$$

Divide both sides by 8a.

$$\frac{8ax}{8a}=\frac{2a\left(a+1\right)}{8a}$$

Dividing by 8a undoes the multiplication by 8a.

$$x=\frac{2a\left(a+1\right)}{8a}$$

Divide 2a\left(1+a\right) by 8a.

$$x=\frac{a+1}{4}$$

Solve for x

$$\left\{\begin{matrix}\\x=\frac{a+1}{4}\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$4x^{2}+4xa+a^{2}-\left(2x-a\right)^{2}-2a=2a^{2}$$

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

$$4x^{2}+4xa+a^{2}-\left(4x^{2}-4xa+a^{2}\right)-2a=2a^{2}$$

To find the opposite of 4x^{2}-4xa+a^{2}, find the opposite of each term.

$$4x^{2}+4xa+a^{2}-4x^{2}+4xa-a^{2}-2a=2a^{2}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$4xa+a^{2}+4xa-a^{2}-2a=2a^{2}$$

Combine 4xa and 4xa to get 8xa.

$$8xa+a^{2}-a^{2}-2a=2a^{2}$$

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

$$8xa-2a=2a^{2}$$

Add 2a to both sides.

$$8xa=2a^{2}+2a$$

The equation is in standard form.

$$8ax=2a^{2}+2a$$

Divide both sides by 8a.

$$\frac{8ax}{8a}=\frac{2a\left(a+1\right)}{8a}$$

Dividing by 8a undoes the multiplication by 8a.

$$x=\frac{2a\left(a+1\right)}{8a}$$

Divide 2a\left(1+a\right) by 8a.

$$x=\frac{a+1}{4}$$

Solve for a

$$a=4x-1$$

$$a=0$$