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6p^2+13p-4=0
a = 6; b = 13; c = -4;
Δ = b2-4ac
Δ = 132-4·6·(-4)
Δ = 265
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{265}}{2*6}=\frac{-13-\sqrt{265}}{12} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{265}}{2*6}=\frac{-13+\sqrt{265}}{12} $
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