2m^2-22m+50=0

Solution for 2m^2-22m+50=0 equation:

2m^2-22m+50=0

a = 2; b = -22; c = +50;
Δ = b2-4ac
Δ = -222-4·2·50
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{21}}{2*2}=\frac{22-2\sqrt{21}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{21}}{2*2}=\frac{22+2\sqrt{21}}{4}
$