2x^2+40x-165=0

Solution for 2x^2+40x-165=0 equation:

2x^2+40x-165=0

a = 2; b = 40; c = -165;
Δ = b2-4ac
Δ = 402-4·2·(-165)
Δ = 2920
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:

$\sqrt{\Delta}=\sqrt{2920}=\sqrt{4*730}=\sqrt{4}*\sqrt{730}=2\sqrt{730}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{730}}{2*2}=\frac{-40-2\sqrt{730}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{730}}{2*2}=\frac{-40+2\sqrt{730}}{4}
$