Solve x^2=(4-x)^2+sqrt3^2

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

Solve for x

$$x = \frac{19}{8} = 2\frac{3}{8} = 2.375$$

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}.

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

The square of \sqrt{3} is 3.

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

Add 16 and 3 to get 19.

$$x^{2}=19-8x+x^{2}$$

Add 8x to both sides.

$$x^{2}+8x=19+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+8x-x^{2}=19$$

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

$$8x=19$$

Divide both sides by 8.

$$x=\frac{19}{8}$$