3y^2-4y-240=0

Solution for 3y^2-4y-240=0 equation:

3y^2-4y-240=0

a = 3; b = -4; c = -240;
Δ = b2-4ac
Δ = -42-4·3·(-240)
Δ = 2896
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:

$\sqrt{\Delta}=\sqrt{2896}=\sqrt{16*181}=\sqrt{16}*\sqrt{181}=4\sqrt{181}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{181}}{2*3}=\frac{4-4\sqrt{181}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{181}}{2*3}=\frac{4+4\sqrt{181}}{6}
$