4x^2+92x-80=0

Solution for 4x^2+92x-80=0 equation:

4x^2+92x-80=0

a = 4; b = 92; c = -80;
Δ = b2-4ac
Δ = 922-4·4·(-80)
Δ = 9744
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:

$\sqrt{\Delta}=\sqrt{9744}=\sqrt{16*609}=\sqrt{16}*\sqrt{609}=4\sqrt{609}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-4\sqrt{609}}{2*4}=\frac{-92-4\sqrt{609}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+4\sqrt{609}}{2*4}=\frac{-92+4\sqrt{609}}{8}
$