Solve (y-6)^2-(y+4)^2=0

$$( y - 6 ) ^ { 2 } - ( y + 4 ) ^ { 2 } = 0$$

Solve for y

$$y=1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-6\right)^{2}.

$$y^{2}-12y+36-\left(y+4\right)^{2}=0$$

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

$$y^{2}-12y+36-\left(y^{2}+8y+16\right)=0$$

To find the opposite of y^{2}+8y+16, find the opposite of each term.

$$y^{2}-12y+36-y^{2}-8y-16=0$$

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

$$-12y+36-8y-16=0$$

Combine -12y and -8y to get -20y.

$$-20y+36-16=0$$

Subtract 16 from 36 to get 20.

$$-20y+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-20y=-20$$

Divide both sides by -20.

$$y=\frac{-20}{-20}$$

Divide -20 by -20 to get 1.

$$y=1$$