-93+n^2-14n=0

Solution for -93+n^2-14n=0 equation:

-93+n^2-14n=0

a = 1; b = -14; c = -93;
Δ = b2-4ac
Δ = -142-4·1·(-93)
Δ = 568
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:

$\sqrt{\Delta}=\sqrt{568}=\sqrt{4*142}=\sqrt{4}*\sqrt{142}=2\sqrt{142}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{142}}{2*1}=\frac{14-2\sqrt{142}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{142}}{2*1}=\frac{14+2\sqrt{142}}{2}
$