16x^2+128x-120=0

Solution for 16x^2+128x-120=0 equation:

16x^2+128x-120=0

a = 16; b = 128; c = -120;
Δ = b2-4ac
Δ = 1282-4·16·(-120)
Δ = 24064
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:

$\sqrt{\Delta}=\sqrt{24064}=\sqrt{256*94}=\sqrt{256}*\sqrt{94}=16\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-16\sqrt{94}}{2*16}=\frac{-128-16\sqrt{94}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+16\sqrt{94}}{2*16}=\frac{-128+16\sqrt{94}}{32}
$