(y+2)(y+2)=2y^2+3y-16

Solution for (y+2)(y+2)=2y^2+3y-16 equation:

(y+2)(y+2)=2y^2+3y-16

We move all terms to the left:

(y+2)(y+2)-(2y^2+3y-16)=0

We get rid of parentheses

-2y^2+(y+2)(y+2)-3y+16=0

We multiply parentheses ..

-2y^2+(+y^2+2y+2y+4)-3y+16=0

We get rid of parentheses

-2y^2+y^2+2y+2y-3y+4+16=0

We add all the numbers together, and all the variables

-1y^2+y+20=0

a = -1; b = 1; c = +20;
Δ = b2-4ac
Δ = 12-4·(-1)·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*-1}=\frac{-10}{-2}
=+5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*-1}=\frac{8}{-2}
=-4
$