20x^2+40x-96=0

Solution for 20x^2+40x-96=0 equation:

20x^2+40x-96=0

a = 20; b = 40; c = -96;
Δ = b2-4ac
Δ = 402-4·20·(-96)
Δ = 9280
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:

$\sqrt{\Delta}=\sqrt{9280}=\sqrt{64*145}=\sqrt{64}*\sqrt{145}=8\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{145}}{2*20}=\frac{-40-8\sqrt{145}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{145}}{2*20}=\frac{-40+8\sqrt{145}}{40}
$