m^2-6m-93=0

Solution for m^2-6m-93=0 equation:

m^2-6m-93=0

a = 1; b = -6; c = -93;
Δ = b2-4ac
Δ = -62-4·1·(-93)
Δ = 408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{102}}{2*1}=\frac{6-2\sqrt{102}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{102}}{2*1}=\frac{6+2\sqrt{102}}{2}
$