n2+9n-22=0

Solution for n2+9n-22=0 equation:

n2+9n-22=0

We add all the numbers together, and all the variables

n^2+9n-22=0

a = 1; b = 9; c = -22;
Δ = b2-4ac
Δ = 92-4·1·(-22)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-13}{2*1}=\frac{-22}{2}
=-11
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+13}{2*1}=\frac{4}{2}
=2
$