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3p^2-28p+9=0
a = 3; b = -28; c = +9;
Δ = b2-4ac
Δ = -282-4·3·9
Δ = 676
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{676}=26$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-26}{2*3}=\frac{2}{6} =1/3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+26}{2*3}=\frac{54}{6} =9 $
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