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6p^2+23p-4=0
a = 6; b = 23; c = -4;
Δ = b2-4ac
Δ = 232-4·6·(-4)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*6}=\frac{-48}{12} =-4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*6}=\frac{2}{12} =1/6 $
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