Solve 2^2-x^2={left(sqrt{13}right)}^2-left(5-xright)^2

$${ 2 }^{ 2 } - { x }^{ 2 } = { \left( \sqrt{ 13 } \right) }^{ 2 } - { \left(5-x \right) }^{ 2 }$$

Solve for x

$$x = \frac{8}{5} = 1\frac{3}{5} = 1.6$$

Steps for Solving Linear Equation

Calculate 2 to the power of 2 and get 4.

$$4-x^{2}=\left(\sqrt{13}\right)^{2}-\left(5-x\right)^{2}$$

The square of \sqrt{13} is 13.

$$4-x^{2}=13-\left(5-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$4-x^{2}=13-\left(25-10x+x^{2}\right)$$

To find the opposite of 25-10x+x^{2}, find the opposite of each term.

$$4-x^{2}=13-25+10x-x^{2}$$

Subtract 25 from 13 to get -12.

$$4-x^{2}=-12+10x-x^{2}$$

Subtract 10x from both sides.

$$4-x^{2}-10x=-12-x^{2}$$

Add x^{2} to both sides.

$$4-x^{2}-10x+x^{2}=-12$$

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

$$4-10x=-12$$

Subtract 4 from both sides.

$$-10x=-12-4$$

Subtract 4 from -12 to get -16.

$$-10x=-16$$

Divide both sides by -10.

$$x=\frac{-16}{-10}$$

Reduce the fraction \frac{-16}{-10} to lowest terms by extracting and canceling out -2.

$$x=\frac{8}{5}$$