n^2+2n-23=0

Solution for n^2+2n-23=0 equation:

n^2+2n-23=0

a = 1; b = 2; c = -23;
Δ = b2-4ac
Δ = 22-4·1·(-23)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{6}}{2*1}=\frac{-2-4\sqrt{6}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{6}}{2*1}=\frac{-2+4\sqrt{6}}{2}
$