2x^2+2x-201=0

Solution for 2x^2+2x-201=0 equation:

2x^2+2x-201=0

a = 2; b = 2; c = -201;
Δ = b2-4ac
Δ = 22-4·2·(-201)
Δ = 1612
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:

$\sqrt{\Delta}=\sqrt{1612}=\sqrt{4*403}=\sqrt{4}*\sqrt{403}=2\sqrt{403}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{403}}{2*2}=\frac{-2-2\sqrt{403}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{403}}{2*2}=\frac{-2+2\sqrt{403}}{4}
$