Solve (y-4)^2+8^2=y^2

$$( y - 4 ) ^ { 2 } + 8 ^ { 2 } = y ^ { 2 }$$

Solve for y

$$y=10$$

Steps for Solving Linear Equation

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

$$y^{2}-8y+16+8^{2}=y^{2}$$

Calculate 8 to the power of 2 and get 64.

$$y^{2}-8y+16+64=y^{2}$$

Add 16 and 64 to get 80.

$$y^{2}-8y+80=y^{2}$$

Subtract y^{2} from both sides.

$$y^{2}-8y+80-y^{2}=0$$

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

$$-8y+80=0$$

Subtract 80 from both sides. Anything subtracted from zero gives its negation.

$$-8y=-80$$

Divide both sides by -8.

$$y=\frac{-80}{-8}$$

Divide -80 by -8 to get 10.

$$y=10$$