Solve 4+x^2={left(2sqrt{5-4}right)}^2+left(4-xright)^2

$$4+ { x }^{ 2 } = { \left(2 \sqrt{ 5-4 } \right) }^{ 2 } + { \left(4-x \right) }^{ 2 }$$

Solve for x

$$x=2$$

Steps for Solving Linear Equation

Subtract 4 from 5 to get 1.

$$4+x^{2}=\left(2\sqrt{1}\right)^{2}+\left(4-x\right)^{2}$$

Calculate the square root of 1 and get 1.

$$4+x^{2}=\left(2\times 1\right)^{2}+\left(4-x\right)^{2}$$

Multiply 2 and 1 to get 2.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Add 4 and 16 to get 20.

$$4+x^{2}=20-8x+x^{2}$$

Add 8x to both sides.

$$4+x^{2}+8x=20+x^{2}$$

Subtract x^{2} from both sides.

$$4+x^{2}+8x-x^{2}=20$$

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

$$4+8x=20$$

Subtract 4 from both sides.

$$8x=20-4$$

Subtract 4 from 20 to get 16.

$$8x=16$$

Divide both sides by 8.

$$x=\frac{16}{8}$$

Divide 16 by 8 to get 2.

$$x=2$$