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6m^2+12m-18=0
a = 6; b = 12; c = -18;
Δ = b2-4ac
Δ = 122-4·6·(-18)
Δ = 576
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{576}=24$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*6}=\frac{-36}{12} =-3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*6}=\frac{12}{12} =1 $
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