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4m^2+3m-45=0
a = 4; b = 3; c = -45;
Δ = b2-4ac
Δ = 32-4·4·(-45)
Δ = 729
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{729}=27$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*4}=\frac{-30}{8} =-3+3/4 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*4}=\frac{24}{8} =3 $
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