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20m^2+55m+30=0
a = 20; b = 55; c = +30;
Δ = b2-4ac
Δ = 552-4·20·30
Δ = 625
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{625}=25$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-25}{2*20}=\frac{-80}{40} =-2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+25}{2*20}=\frac{-30}{40} =-3/4 $
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