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

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

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps for Solving Linear Equation

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

$$x^{2}=16-8x+x^{2}+\left(2\sqrt{3}\right)^{2}$$

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

$$x^{2}=16-8x+x^{2}+2^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}=16-8x+x^{2}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$x^{2}=16-8x+x^{2}+4\times 3$$

Multiply 4 and 3 to get 12.

$$x^{2}=16-8x+x^{2}+12$$

Add 16 and 12 to get 28.

$$x^{2}=28-8x+x^{2}$$

Add 8x to both sides.

$$x^{2}+8x=28+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+8x-x^{2}=28$$

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

$$8x=28$$

Divide both sides by 8.

$$x=\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{7}{2}$$