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 = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{70}}{2*2}=\frac{-40-2\sqrt{70}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{70}}{2*2}=\frac{-40+2\sqrt{70}}{4}
$