2x^2+20x-9900=0

Solution for 2x^2+20x-9900=0 equation:

2x^2+20x-9900=0

a = 2; b = 20; c = -9900;
Δ = b2-4ac
Δ = 202-4·2·(-9900)
Δ = 79600
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:

$\sqrt{\Delta}=\sqrt{79600}=\sqrt{400*199}=\sqrt{400}*\sqrt{199}=20\sqrt{199}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{199}}{2*2}=\frac{-20-20\sqrt{199}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{199}}{2*2}=\frac{-20+20\sqrt{199}}{4}
$