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3m^2-25m-18=0
a = 3; b = -25; c = -18;
Δ = b2-4ac
Δ = -252-4·3·(-18)
Δ = 841
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{841}=29$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-29}{2*3}=\frac{-4}{6} =-2/3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+29}{2*3}=\frac{54}{6} =9 $
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