5m^2-50m+98=0

Solution for 5m^2-50m+98=0 equation:

5m^2-50m+98=0

a = 5; b = -50; c = +98;
Δ = b2-4ac
Δ = -502-4·5·98
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-6\sqrt{15}}{2*5}=\frac{50-6\sqrt{15}}{10}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+6\sqrt{15}}{2*5}=\frac{50+6\sqrt{15}}{10}
$