m^2-14m+19=0

Solution for m^2-14m+19=0 equation:

m^2-14m+19=0

a = 1; b = -14; c = +19;
Δ = b2-4ac
Δ = -142-4·1·19
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{30}}{2*1}=\frac{14-2\sqrt{30}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{30}}{2*1}=\frac{14+2\sqrt{30}}{2}
$