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15p^2+13p+2=0
a = 15; b = 13; c = +2;
Δ = b2-4ac
Δ = 132-4·15·2
Δ = 49
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{49}=7$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*15}=\frac{-20}{30} =-2/3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*15}=\frac{-6}{30} =-1/5 $
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