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-3p^2+11p-4=0
a = -3; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·(-3)·(-4)
Δ = 73
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{-(11)-\sqrt{73}}{2*-3}=\frac{-11-\sqrt{73}}{-6} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{73}}{2*-3}=\frac{-11+\sqrt{73}}{-6} $
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