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2m^2-19m+24=0
a = 2; b = -19; c = +24;
Δ = b2-4ac
Δ = -192-4·2·24
Δ = 169
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{169}=13$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*2}=\frac{6}{4} =1+1/2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*2}=\frac{32}{4} =8 $
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