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6m^2+23m-18=0
a = 6; b = 23; c = -18;
Δ = b2-4ac
Δ = 232-4·6·(-18)
Δ = 961
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{961}=31$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-31}{2*6}=\frac{-54}{12} =-4+1/2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+31}{2*6}=\frac{8}{12} =2/3 $
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