Solve {l}{2^2+x^2=(3-x)^2+1*6^2}{=}

$$\left. \begin{array}{l}{ 2 ^ { 2 } + x ^ { 2 } = ( 3 - x ) ^ { 2 } + 1 \cdot 6 ^ { 2 } }\\{ = }\end{array} \right.$$

Solve for x

$$x = \frac{41}{6} = 6\frac{5}{6} \approx 6.833333333$$

Steps for Solving Linear Equation

Calculate 2 to the power of 2 and get 4.

$$4+x^{2}=\left(3-x\right)^{2}+1\times 6^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$4+x^{2}=9-6x+x^{2}+1\times 6^{2}$$

Calculate 6 to the power of 2 and get 36.

$$4+x^{2}=9-6x+x^{2}+1\times 36$$

Multiply 1 and 36 to get 36.

$$4+x^{2}=9-6x+x^{2}+36$$

Add 9 and 36 to get 45.

$$4+x^{2}=45-6x+x^{2}$$

Add 6x to both sides.

$$4+x^{2}+6x=45+x^{2}$$

Subtract x^{2} from both sides.

$$4+x^{2}+6x-x^{2}=45$$

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

$$4+6x=45$$

Subtract 4 from both sides.

$$6x=45-4$$

Subtract 4 from 45 to get 41.

$$6x=41$$

Divide both sides by 6.

$$x=\frac{41}{6}$$