Solve {left(5/2+3right)}^2+x^2={left(5/2+5right)}^2+left(4-xright)^2

$${ \left( \frac{ 5 }{ 2 } +3 \right) }^{ 2 } + { x }^{ 2 } = { \left( \frac{ 5 }{ 2 } +5 \right) }^{ 2 } + { \left(4-x \right) }^{ 2 }$$

Solve for x

$$x = \frac{21}{4} = 5\frac{1}{4} = 5.25$$

Steps for Solving Linear Equation

Add \frac{5}{2} and 3 to get \frac{11}{2}.

$$\left(\frac{11}{2}\right)^{2}+x^{2}=\left(\frac{5}{2}+5\right)^{2}+\left(4-x\right)^{2}$$

Calculate \frac{11}{2} to the power of 2 and get \frac{121}{4}.

$$\frac{121}{4}+x^{2}=\left(\frac{5}{2}+5\right)^{2}+\left(4-x\right)^{2}$$

Add \frac{5}{2} and 5 to get \frac{15}{2}.

$$\frac{121}{4}+x^{2}=\left(\frac{15}{2}\right)^{2}+\left(4-x\right)^{2}$$

Calculate \frac{15}{2} to the power of 2 and get \frac{225}{4}.

$$\frac{121}{4}+x^{2}=\frac{225}{4}+\left(4-x\right)^{2}$$

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

$$\frac{121}{4}+x^{2}=\frac{225}{4}+16-8x+x^{2}$$

Add \frac{225}{4} and 16 to get \frac{289}{4}.

$$\frac{121}{4}+x^{2}=\frac{289}{4}-8x+x^{2}$$

Add 8x to both sides.

$$\frac{121}{4}+x^{2}+8x=\frac{289}{4}+x^{2}$$

Subtract x^{2} from both sides.

$$\frac{121}{4}+x^{2}+8x-x^{2}=\frac{289}{4}$$

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

$$\frac{121}{4}+8x=\frac{289}{4}$$

Subtract \frac{121}{4} from both sides.

$$8x=\frac{289}{4}-\frac{121}{4}$$

Subtract \frac{121}{4} from \frac{289}{4} to get 42.

$$8x=42$$

Divide both sides by 8.

$$x=\frac{42}{8}$$

Reduce the fraction \frac{42}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{21}{4}$$