n^2+24n-116=0

Solution for n^2+24n-116=0 equation:

n^2+24n-116=0

a = 1; b = 24; c = -116;
Δ = b2-4ac
Δ = 242-4·1·(-116)
Δ = 1040
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1040}=\sqrt{16*65}=\sqrt{16}*\sqrt{65}=4\sqrt{65}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{65}}{2*1}=\frac{-24-4\sqrt{65}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{65}}{2*1}=\frac{-24+4\sqrt{65}}{2}
$