20x^2+8x-48=0

Solution for 20x^2+8x-48=0 equation:

20x^2+8x-48=0

a = 20; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·20·(-48)
Δ = 3904
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:

$\sqrt{\Delta}=\sqrt{3904}=\sqrt{64*61}=\sqrt{64}*\sqrt{61}=8\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{61}}{2*20}=\frac{-8-8\sqrt{61}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{61}}{2*20}=\frac{-8+8\sqrt{61}}{40}
$