5x^2+20x-16=0

Solution for 5x^2+20x-16=0 equation:

5x^2+20x-16=0

a = 5; b = 20; c = -16;
Δ = b2-4ac
Δ = 202-4·5·(-16)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{5}}{2*5}=\frac{-20-12\sqrt{5}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{5}}{2*5}=\frac{-20+12\sqrt{5}}{10}
$