2x^2+3x-540=0

Solution for 2x^2+3x-540=0 equation:

2x^2+3x-540=0

a = 2; b = 3; c = -540;
Δ = b2-4ac
Δ = 32-4·2·(-540)
Δ = 4329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4329}=\sqrt{9*481}=\sqrt{9}*\sqrt{481}=3\sqrt{481}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{481}}{2*2}=\frac{-3-3\sqrt{481}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{481}}{2*2}=\frac{-3+3\sqrt{481}}{4}
$