x^2+24x-160=0

Solution for x^2+24x-160=0 equation:

x^2+24x-160=0

a = 1; b = 24; c = -160;
Δ = b2-4ac
Δ = 242-4·1·(-160)
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{19}}{2*1}=\frac{-24-8\sqrt{19}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{19}}{2*1}=\frac{-24+8\sqrt{19}}{2}
$