Solve left(x-1.8right)^2+2.4^2=x^2

$${ \left(x-1.8 \right) }^{ 2 } + { 2.4 }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x=2.5$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1.8\right)^{2}.

$$x^{2}-3.6x+3.24+2.4^{2}=x^{2}$$

Calculate 2.4 to the power of 2 and get 5.76.

$$x^{2}-3.6x+3.24+5.76=x^{2}$$

Add 3.24 and 5.76 to get 9.

$$x^{2}-3.6x+9=x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-3.6x+9-x^{2}=0$$

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

$$-3.6x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-3.6x=-9$$

Divide both sides by -3.6.

$$x=\frac{-9}{-3.6}$$

Expand \frac{-9}{-3.6} by multiplying both numerator and the denominator by 10.

$$x=\frac{-90}{-36}$$

Reduce the fraction \frac{-90}{-36} to lowest terms by extracting and canceling out -18.

$$x=\frac{5}{2}$$