4x^2+20x-384=0

Solution for 4x^2+20x-384=0 equation:

4x^2+20x-384=0

a = 4; b = 20; c = -384;
Δ = b2-4ac
Δ = 202-4·4·(-384)
Δ = 6544
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:

$\sqrt{\Delta}=\sqrt{6544}=\sqrt{16*409}=\sqrt{16}*\sqrt{409}=4\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{409}}{2*4}=\frac{-20-4\sqrt{409}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{409}}{2*4}=\frac{-20+4\sqrt{409}}{8}
$