20x^2+40x-200=0

Solution for 20x^2+40x-200=0 equation:

20x^2+40x-200=0

a = 20; b = 40; c = -200;
Δ = b2-4ac
Δ = 402-4·20·(-200)
Δ = 17600
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:

$\sqrt{\Delta}=\sqrt{17600}=\sqrt{1600*11}=\sqrt{1600}*\sqrt{11}=40\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{11}}{2*20}=\frac{-40-40\sqrt{11}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{11}}{2*20}=\frac{-40+40\sqrt{11}}{40}
$