2y^2+15y-495=0

Solution for 2y^2+15y-495=0 equation:

2y^2+15y-495=0

a = 2; b = 15; c = -495;
Δ = b2-4ac
Δ = 152-4·2·(-495)
Δ = 4185
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:

$\sqrt{\Delta}=\sqrt{4185}=\sqrt{9*465}=\sqrt{9}*\sqrt{465}=3\sqrt{465}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{465}}{2*2}=\frac{-15-3\sqrt{465}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{465}}{2*2}=\frac{-15+3\sqrt{465}}{4}
$