20x^2+48x-19=0

Solution for 20x^2+48x-19=0 equation:

20x^2+48x-19=0

a = 20; b = 48; c = -19;
Δ = b2-4ac
Δ = 482-4·20·(-19)
Δ = 3824
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:

$\sqrt{\Delta}=\sqrt{3824}=\sqrt{16*239}=\sqrt{16}*\sqrt{239}=4\sqrt{239}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{239}}{2*20}=\frac{-48-4\sqrt{239}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{239}}{2*20}=\frac{-48+4\sqrt{239}}{40}
$