5m^2-21m+22=0

Solution for 5m^2-21m+22=0 equation:

5m^2-21m+22=0

a = 5; b = -21; c = +22;
Δ = b2-4ac
Δ = -212-4·5·22
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-1}{2*5}=\frac{20}{10}
=2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+1}{2*5}=\frac{22}{10}
=2+1/5
$