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6p^2+33p-120=0
a = 6; b = 33; c = -120;
Δ = b2-4ac
Δ = 332-4·6·(-120)
Δ = 3969
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{3969}=63$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-63}{2*6}=\frac{-96}{12} =-8 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+63}{2*6}=\frac{30}{12} =2+1/2 $
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