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2m^2+3m-35=0
a = 2; b = 3; c = -35;
Δ = b2-4ac
Δ = 32-4·2·(-35)
Δ = 289
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{289}=17$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-17}{2*2}=\frac{-20}{4} =-5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+17}{2*2}=\frac{14}{4} =3+1/2 $
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