Solve (x-0.8)^2+2.4^2=x^2

$$( x - 0.8 ) ^ { 2 } + 2.4 ^ { 2 } = x ^ { 2 }$$

Solve for x

$$x=4$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-0.8\right)^{2}.

$$x^{2}-1.6x+0.64+2.4^{2}=x^{2}$$

Calculate 2.4 to the power of 2 and get 5.76.

$$x^{2}-1.6x+0.64+5.76=x^{2}$$

Add 0.64 and 5.76 to get 6.4.

$$x^{2}-1.6x+6.4=x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-1.6x+6.4-x^{2}=0$$

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

$$-1.6x+6.4=0$$

Subtract 6.4 from both sides. Anything subtracted from zero gives its negation.

$$-1.6x=-6.4$$

Divide both sides by -1.6.

$$x=\frac{-6.4}{-1.6}$$

Expand \frac{-6.4}{-1.6} by multiplying both numerator and the denominator by 10.

$$x=\frac{-64}{-16}$$

Divide -64 by -16 to get 4.

$$x=4$$