Solve (2a+3b)^2=4a^2+1

$$( 2 a + 3 b ) ^ { 2 } = 4 a ^ { 2 } + 1$$

Solve for a

$$a=-\frac{3b}{4}+\frac{1}{12b}$$

$$b\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2a+3b\right)^{2}.

$$4a^{2}+12ab+9b^{2}=4a^{2}+1$$

Subtract 4a^{2} from both sides.

$$4a^{2}+12ab+9b^{2}-4a^{2}=1$$

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

$$12ab+9b^{2}=1$$

Subtract 9b^{2} from both sides.

$$12ab=1-9b^{2}$$

The equation is in standard form.

$$12ba=1-9b^{2}$$

Divide both sides by 12b.

$$\frac{12ba}{12b}=\frac{1-9b^{2}}{12b}$$

Dividing by 12b undoes the multiplication by 12b.

$$a=\frac{1-9b^{2}}{12b}$$

Divide -9b^{2}+1 by 12b.

$$a=-\frac{3b}{4}+\frac{1}{12b}$$

Solve for b

$$b=\frac{\sqrt{4a^{2}+1}-2a}{3}$$

$$b=\frac{-\sqrt{4a^{2}+1}-2a}{3}$$