n^2+2n-22=0

Solution for n^2+2n-22=0 equation:

n^2+2n-22=0

a = 1; b = 2; c = -22;
Δ = b2-4ac
Δ = 22-4·1·(-22)
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{23}}{2*1}=\frac{-2-2\sqrt{23}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{23}}{2*1}=\frac{-2+2\sqrt{23}}{2}
$