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2m^2-3m+1=0
a = 2; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·2·1
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*2}=\frac{2}{4} =1/2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*2}=\frac{4}{4} =1 $
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