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3m^2+2m-65=0
a = 3; b = 2; c = -65;
Δ = b2-4ac
Δ = 22-4·3·(-65)
Δ = 784
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{784}=28$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-28}{2*3}=\frac{-30}{6} =-5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+28}{2*3}=\frac{26}{6} =4+1/3 $
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