(3y-2)(y+4)-24=0

Solution for (3y-2)(y+4)-24=0 equation:

(3y-2)(y+4)-24=0

We multiply parentheses ..

(+3y^2+12y-2y-8)-24=0

We get rid of parentheses

3y^2+12y-2y-8-24=0

We add all the numbers together, and all the variables

3y^2+10y-32=0

a = 3; b = 10; c = -32;
Δ = b2-4ac
Δ = 102-4·3·(-32)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*3}=\frac{-32}{6}
=-5+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*3}=\frac{12}{6}
=2
$