80x^2+160x-12=0

Solution for 80x^2+160x-12=0 equation:

80x^2+160x-12=0

a = 80; b = 160; c = -12;
Δ = b2-4ac
Δ = 1602-4·80·(-12)
Δ = 29440
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:

$\sqrt{\Delta}=\sqrt{29440}=\sqrt{256*115}=\sqrt{256}*\sqrt{115}=16\sqrt{115}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-16\sqrt{115}}{2*80}=\frac{-160-16\sqrt{115}}{160}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+16\sqrt{115}}{2*80}=\frac{-160+16\sqrt{115}}{160}
$