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4m^2-13m+10=0
a = 4; b = -13; c = +10;
Δ = b2-4ac
Δ = -132-4·4·10
Δ = 9
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{9}=3$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*4}=\frac{10}{8} =1+1/4 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*4}=\frac{16}{8} =2 $
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