Solve (4sqrt{3})^2+x^2=(12-x)^2

$$( 4 \sqrt { 3 } ) ^ { 2 } + x ^ { 2 } = ( 12 - x ) ^ { 2 }$$

Solve for x

$$x=4$$

Steps for Solving Linear Equation

Expand \left(4\sqrt{3}\right)^{2}.

$$4^{2}\left(\sqrt{3}\right)^{2}+x^{2}=\left(12-x\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16\left(\sqrt{3}\right)^{2}+x^{2}=\left(12-x\right)^{2}$$

The square of \sqrt{3} is 3.

$$16\times 3+x^{2}=\left(12-x\right)^{2}$$

Multiply 16 and 3 to get 48.

$$48+x^{2}=\left(12-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-x\right)^{2}.

$$48+x^{2}=144-24x+x^{2}$$

Add 24x to both sides.

$$48+x^{2}+24x=144+x^{2}$$

Subtract x^{2} from both sides.

$$48+x^{2}+24x-x^{2}=144$$

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

$$48+24x=144$$

Subtract 48 from both sides.

$$24x=144-48$$

Subtract 48 from 144 to get 96.

$$24x=96$$

Divide both sides by 24.

$$x=\frac{96}{24}$$

Divide 96 by 24 to get 4.

$$x=4$$