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3p^2-14p+15=0
a = 3; b = -14; c = +15;
Δ = b2-4ac
Δ = -142-4·3·15
Δ = 16
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{16}=4$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*3}=\frac{10}{6} =1+2/3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*3}=\frac{18}{6} =3 $
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