16x^2+32x-120=0

Solution for 16x^2+32x-120=0 equation:

16x^2+32x-120=0

a = 16; b = 32; c = -120;
Δ = b2-4ac
Δ = 322-4·16·(-120)
Δ = 8704
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:

$\sqrt{\Delta}=\sqrt{8704}=\sqrt{256*34}=\sqrt{256}*\sqrt{34}=16\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{34}}{2*16}=\frac{-32-16\sqrt{34}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{34}}{2*16}=\frac{-32+16\sqrt{34}}{32}
$