4x^2+16x-80=0

Solution for 4x^2+16x-80=0 equation:

4x^2+16x-80=0

a = 4; b = 16; c = -80;
Δ = b2-4ac
Δ = 162-4·4·(-80)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{6}}{2*4}=\frac{-16-16\sqrt{6}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{6}}{2*4}=\frac{-16+16\sqrt{6}}{8}
$