(y+1)(y-1)=-8y-13

Solution for (y+1)(y-1)=-8y-13 equation:

(y+1)(y-1)=-8y-13

We move all terms to the left:

(y+1)(y-1)-(-8y-13)=0

We use the square of the difference formula

y^2-(-8y-13)-1=0

We get rid of parentheses

y^2+8y+13-1=0

We add all the numbers together, and all the variables

y^2+8y+12=0

a = 1; b = 8; c = +12;
Δ = b2-4ac
Δ = 82-4·1·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*1}=\frac{-12}{2}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*1}=\frac{-4}{2}
=-2
$