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15m^2+23m+6=0
a = 15; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·15·6
Δ = 169
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{169}=13$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-13}{2*15}=\frac{-36}{30} =-1+1/5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+13}{2*15}=\frac{-10}{30} =-1/3 $
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