Solve (2sqrt{3})^2-a^2=2^2-(4-a)^2

$$( 2 \sqrt { 3 } ) ^ { 2 } - a ^ { 2 } = 2 ^ { 2 } - ( 4 - a ) ^ { 2 }$$

Solve for a

$$a=3$$

Steps for Solving Linear Equation

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}-a^{2}=2^{2}-\left(4-a\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}-a^{2}=2^{2}-\left(4-a\right)^{2}$$

The square of \sqrt{3} is 3.

$$4\times 3-a^{2}=2^{2}-\left(4-a\right)^{2}$$

Multiply 4 and 3 to get 12.

$$12-a^{2}=2^{2}-\left(4-a\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$12-a^{2}=4-\left(4-a\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-a\right)^{2}.

$$12-a^{2}=4-\left(16-8a+a^{2}\right)$$

To find the opposite of 16-8a+a^{2}, find the opposite of each term.

$$12-a^{2}=4-16+8a-a^{2}$$

Subtract 16 from 4 to get -12.

$$12-a^{2}=-12+8a-a^{2}$$

Subtract 8a from both sides.

$$12-a^{2}-8a=-12-a^{2}$$

Add a^{2} to both sides.

$$12-a^{2}-8a+a^{2}=-12$$

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

$$12-8a=-12$$

Subtract 12 from both sides.

$$-8a=-12-12$$

Subtract 12 from -12 to get -24.

$$-8a=-24$$

Divide both sides by -8.

$$a=\frac{-24}{-8}$$

Divide -24 by -8 to get 3.

$$a=3$$