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3m^2+3m-90=0
a = 3; b = 3; c = -90;
Δ = b2-4ac
Δ = 32-4·3·(-90)
Δ = 1089
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{1089}=33$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-33}{2*3}=\frac{-36}{6} =-6 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+33}{2*3}=\frac{30}{6} =5 $
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