n^2+14n-91=0

Solution for n^2+14n-91=0 equation:

n^2+14n-91=0

a = 1; b = 14; c = -91;
Δ = b2-4ac
Δ = 142-4·1·(-91)
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{35}}{2*1}=\frac{-14-4\sqrt{35}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{35}}{2*1}=\frac{-14+4\sqrt{35}}{2}
$