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2m^2-3m-44=0
a = 2; b = -3; c = -44;
Δ = b2-4ac
Δ = -32-4·2·(-44)
Δ = 361
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{361}=19$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-19}{2*2}=\frac{-16}{4} =-4 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+19}{2*2}=\frac{22}{4} =5+1/2 $
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