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3m^2+4m-15=0
a = 3; b = 4; c = -15;
Δ = b2-4ac
Δ = 42-4·3·(-15)
Δ = 196
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{196}=14$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*3}=\frac{-18}{6} =-3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*3}=\frac{10}{6} =1+2/3 $
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