Solve (y+4)^2=(y-3)^2

$$( y + 4 ) ^ { 2 } = ( y - 3 ) ^ { 2 }$$

Solve for y

$$y=-\frac{1}{2}=-0.5$$

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}.

$$y^{2}+8y+16=\left(y-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$y^{2}+8y+16=y^{2}-6y+9$$

Subtract y^{2} from both sides.

$$y^{2}+8y+16-y^{2}=-6y+9$$

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

$$8y+16=-6y+9$$

Add 6y to both sides.

$$8y+16+6y=9$$

Combine 8y and 6y to get 14y.

$$14y+16=9$$

Subtract 16 from both sides.

$$14y=9-16$$

Subtract 16 from 9 to get -7.

$$14y=-7$$

Divide both sides by 14.

$$y=\frac{-7}{14}$$

Reduce the fraction \frac{-7}{14} to lowest terms by extracting and canceling out 7.

$$y=-\frac{1}{2}$$