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21m^2-13m+2=0
a = 21; b = -13; c = +2;
Δ = b2-4ac
Δ = -132-4·21·2
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-1}{2*21}=\frac{12}{42} =2/7 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+1}{2*21}=\frac{14}{42} =1/3 $
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