Solve left(x-0.6right)^2+4.2^2=x^2

$${ \left(x-0.6 \right) }^{ 2 } + { 4.2 }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x=15$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-0.6\right)^{2}.

$$x^{2}-1.2x+0.36+4.2^{2}=x^{2}$$

Calculate 4.2 to the power of 2 and get 17.64.

$$x^{2}-1.2x+0.36+17.64=x^{2}$$

Add 0.36 and 17.64 to get 18.

$$x^{2}-1.2x+18=x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-1.2x+18-x^{2}=0$$

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

$$-1.2x+18=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$-1.2x=-18$$

Divide both sides by -1.2.

$$x=\frac{-18}{-1.2}$$

Expand \frac{-18}{-1.2} by multiplying both numerator and the denominator by 10.

$$x=\frac{-180}{-12}$$

Divide -180 by -12 to get 15.

$$x=15$$