Solve left(4-xright)^2=x^2+2.5^2

$${ \left(4-x \right) }^{ 2 } = { x }^{ 2 } + { 2.5 }^{ 2 }$$

Solve for x

$$x=1.21875$$

Steps for Solving Linear Equation

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

$$16-8x+x^{2}=x^{2}+2.5^{2}$$

Calculate 2.5 to the power of 2 and get 6.25.

$$16-8x+x^{2}=x^{2}+6.25$$

Subtract x^{2} from both sides.

$$16-8x+x^{2}-x^{2}=6.25$$

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

$$16-8x=6.25$$

Subtract 16 from both sides.

$$-8x=6.25-16$$

Subtract 16 from 6.25 to get -9.75.

$$-8x=-9.75$$

Divide both sides by -8.

$$x=\frac{-9.75}{-8}$$

Expand \frac{-9.75}{-8} by multiplying both numerator and the denominator by 100.

$$x=\frac{-975}{-800}$$

Reduce the fraction \frac{-975}{-800} to lowest terms by extracting and canceling out -25.

$$x=\frac{39}{32}$$