20x^2+56x-96=0

Solution for 20x^2+56x-96=0 equation:

20x^2+56x-96=0

a = 20; b = 56; c = -96;
Δ = b2-4ac
Δ = 562-4·20·(-96)
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-104}{2*20}=\frac{-160}{40}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+104}{2*20}=\frac{48}{40}
=1+1/5
$