20x^2-28x-96=0

Solution for 20x^2-28x-96=0 equation:

20x^2-28x-96=0

a = 20; b = -28; c = -96;
Δ = b2-4ac
Δ = -282-4·20·(-96)
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-92}{2*20}=\frac{-64}{40}
=-1+3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+92}{2*20}=\frac{120}{40}
=3
$