m^2-14m+22=0

Solution for m^2-14m+22=0 equation:

m^2-14m+22=0

a = 1; b = -14; c = +22;
Δ = b2-4ac
Δ = -142-4·1·22
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6\sqrt{3}}{2*1}=\frac{14-6\sqrt{3}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6\sqrt{3}}{2*1}=\frac{14+6\sqrt{3}}{2}
$