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20p^2+33p+16=6
We move all terms to the left:
20p^2+33p+16-(6)=0
We add all the numbers together, and all the variables
20p^2+33p+10=0
a = 20; b = 33; c = +10;
Δ = b2-4ac
Δ = 332-4·20·10
Δ = 289
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{289}=17$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-17}{2*20}=\frac{-50}{40} =-1+1/4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+17}{2*20}=\frac{-16}{40} =-2/5 $
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