2x^2+24x-280=0

Solution for 2x^2+24x-280=0 equation:

2x^2+24x-280=0

a = 2; b = 24; c = -280;
Δ = b2-4ac
Δ = 242-4·2·(-280)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16\sqrt{11}}{2*2}=\frac{-24-16\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16\sqrt{11}}{2*2}=\frac{-24+16\sqrt{11}}{4}
$