-3y^2+20y+32=0

Solution for -3y^2+20y+32=0 equation:

-3y^2+20y+32=0

a = -3; b = 20; c = +32;
Δ = b2-4ac
Δ = 202-4·(-3)·32
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*-3}=\frac{-48}{-6}
=+8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*-3}=\frac{8}{-6}
=-1+1/3
$