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6p^2-33p+16=0
a = 6; b = -33; c = +16;
Δ = b2-4ac
Δ = -332-4·6·16
Δ = 705
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{-(-33)-\sqrt{705}}{2*6}=\frac{33-\sqrt{705}}{12} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{705}}{2*6}=\frac{33+\sqrt{705}}{12} $
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