n^2+14n-91=5

Solution for n^2+14n-91=5 equation:

n^2+14n-91=5

We move all terms to the left:

n^2+14n-91-(5)=0

We add all the numbers together, and all the variables

n^2+14n-96=0

a = 1; b = 14; c = -96;
Δ = b2-4ac
Δ = 142-4·1·(-96)
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{145}}{2*1}=\frac{-14-2\sqrt{145}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{145}}{2*1}=\frac{-14+2\sqrt{145}}{2}
$