n2+4n-63=5

Solution for n2+4n-63=5 equation:

n2+4n-63=5

We move all terms to the left:

n2+4n-63-(5)=0

We add all the numbers together, and all the variables

n^2+4n-68=0

a = 1; b = 4; c = -68;
Δ = b2-4ac
Δ = 42-4·1·(-68)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12\sqrt{2}}{2*1}=\frac{-4-12\sqrt{2}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12\sqrt{2}}{2*1}=\frac{-4+12\sqrt{2}}{2}
$