Solve (2a+b)^2=4a^2+4a

$$( 2 a + b ) ^ { 2 } = 4 a ^ { 2 } + 4 a$$

Solve for a

$$a=-\frac{b^{2}}{4\left(b-1\right)}$$

$$b\neq 1$$

Steps for Solving Linear Equation

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

$$4a^{2}+4ab+b^{2}=4a^{2}+4a$$

Subtract 4a^{2} from both sides.

$$4a^{2}+4ab+b^{2}-4a^{2}=4a$$

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

$$4ab+b^{2}=4a$$

Subtract 4a from both sides.

$$4ab+b^{2}-4a=0$$

Subtract b^{2} from both sides. Anything subtracted from zero gives its negation.

$$4ab-4a=-b^{2}$$

Combine all terms containing a.

$$\left(4b-4\right)a=-b^{2}$$

Divide both sides by 4b-4.

$$\frac{\left(4b-4\right)a}{4b-4}=-\frac{b^{2}}{4b-4}$$

Dividing by 4b-4 undoes the multiplication by 4b-4.

$$a=-\frac{b^{2}}{4b-4}$$

Divide -b^{2} by 4b-4.

$$a=-\frac{b^{2}}{4\left(b-1\right)}$$

Solve for b (complex solution)

$$b=-2\sqrt{a}\sqrt{a+1}-2a$$

$$b=2\sqrt{a}\sqrt{a+1}-2a$$

Solve for b

$$b=-2\sqrt{a^{2}+a}-2a$$

$$b=2\sqrt{a^{2}+a}-2a\text{, }a\leq -1\text{ or }a\geq 0$$