2x^2+24x-20=0

Solution for 2x^2+24x-20=0 equation:

2x^2+24x-20=0

a = 2; b = 24; c = -20;
Δ = b2-4ac
Δ = 242-4·2·(-20)
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{46}}{2*2}=\frac{-24-4\sqrt{46}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{46}}{2*2}=\frac{-24+4\sqrt{46}}{4}
$