n^2+12n-53=-9

Solution for n^2+12n-53=-9 equation:

n^2+12n-53=-9

We move all terms to the left:

n^2+12n-53-(-9)=0

We add all the numbers together, and all the variables

n^2+12n-44=0

a = 1; b = 12; c = -44;
Δ = b2-4ac
Δ = 122-4·1·(-44)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{5}}{2*1}=\frac{-12-8\sqrt{5}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{5}}{2*1}=\frac{-12+8\sqrt{5}}{2}
$