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3m^2-6m-24=0
a = 3; b = -6; c = -24;
Δ = b2-4ac
Δ = -62-4·3·(-24)
Δ = 324
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{324}=18$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*3}=\frac{-12}{6} =-2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*3}=\frac{24}{6} =4 $
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