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3p^2-14p=0
a = 3; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·3·0
Δ = 196
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{196}=14$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*3}=\frac{0}{6} =0 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*3}=\frac{28}{6} =4+2/3 $
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