20x^2+112x-48=0

Solution for 20x^2+112x-48=0 equation:

20x^2+112x-48=0

a = 20; b = 112; c = -48;
Δ = b2-4ac
Δ = 1122-4·20·(-48)
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-128}{2*20}=\frac{-240}{40}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+128}{2*20}=\frac{16}{40}
=2/5
$