n^2+2n-28=-9

Solution for n^2+2n-28=-9 equation:

n^2+2n-28=-9

We move all terms to the left:

n^2+2n-28-(-9)=0

We add all the numbers together, and all the variables

n^2+2n-19=0

a = 1; b = 2; c = -19;
Δ = b2-4ac
Δ = 22-4·1·(-19)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{5}}{2*1}=\frac{-2-4\sqrt{5}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{5}}{2*1}=\frac{-2+4\sqrt{5}}{2}
$