3y^2+8y-120=0

Solution for 3y^2+8y-120=0 equation:

3y^2+8y-120=0

a = 3; b = 8; c = -120;
Δ = b2-4ac
Δ = 82-4·3·(-120)
Δ = 1504
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:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{94}}{2*3}=\frac{-8-4\sqrt{94}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{94}}{2*3}=\frac{-8+4\sqrt{94}}{6}
$