Solve {left(2sqrt{13}right)}^2-left(5-+xright)^2=9-x^2

$${ \left(2 \sqrt{ 13 } \right) }^{ 2 } - { \left(5-+x \right) }^{ 2 } =9- { x }^{ 2 }$$

Solve for x

$$x = -\frac{9}{5} = -1\frac{4}{5} = -1.8$$

Steps for Solving Linear Equation

Expand \left(2\sqrt{13}\right)^{2}.

$$2^{2}\left(\sqrt{13}\right)^{2}-\left(5-x\right)^{2}=9-x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{13}\right)^{2}-\left(5-x\right)^{2}=9-x^{2}$$

The square of \sqrt{13} is 13.

$$4\times 13-\left(5-x\right)^{2}=9-x^{2}$$

Multiply 4 and 13 to get 52.

$$52-\left(5-x\right)^{2}=9-x^{2}$$

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

$$52-\left(25-10x+x^{2}\right)=9-x^{2}$$

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

$$52-25+10x-x^{2}=9-x^{2}$$

Subtract 25 from 52 to get 27.

$$27+10x-x^{2}=9-x^{2}$$

Add x^{2} to both sides.

$$27+10x-x^{2}+x^{2}=9$$

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

$$27+10x=9$$

Subtract 27 from both sides.

$$10x=9-27$$

Subtract 27 from 9 to get -18.

$$10x=-18$$

Divide both sides by 10.

$$x=\frac{-18}{10}$$

Reduce the fraction \frac{-18}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{9}{5}$$