(4y+4)(y+1)=324

Solution for (4y+4)(y+1)=324 equation:

(4y+4)(y+1)=324

We move all terms to the left:

(4y+4)(y+1)-(324)=0

We multiply parentheses ..

(+4y^2+4y+4y+4)-324=0

We get rid of parentheses

4y^2+4y+4y+4-324=0

We add all the numbers together, and all the variables

4y^2+8y-320=0

a = 4; b = 8; c = -320;
Δ = b2-4ac
Δ = 82-4·4·(-320)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-72}{2*4}=\frac{-80}{8}
=-10
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+72}{2*4}=\frac{64}{8}
=8
$