3y^2+12y-32=4

Solution for 3y^2+12y-32=4 equation:

3y^2+12y-32=4

We move all terms to the left:

3y^2+12y-32-(4)=0

We add all the numbers together, and all the variables

3y^2+12y-36=0

a = 3; b = 12; c = -36;
Δ = b2-4ac
Δ = 122-4·3·(-36)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*3}=\frac{-36}{6}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*3}=\frac{12}{6}
=2
$