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18p^2+33p+15=0
a = 18; b = 33; c = +15;
Δ = b2-4ac
Δ = 332-4·18·15
Δ = 9
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{9}=3$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3}{2*18}=\frac{-36}{36} =-1 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3}{2*18}=\frac{-30}{36} =-5/6 $
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