3y^2-20y-12=0

Solution for 3y^2-20y-12=0 equation:

3y^2-20y-12=0

a = 3; b = -20; c = -12;
Δ = b2-4ac
Δ = -202-4·3·(-12)
Δ = 544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{34}}{2*3}=\frac{20-4\sqrt{34}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{34}}{2*3}=\frac{20+4\sqrt{34}}{6}
$