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p2-33p+16=0
We add all the numbers together, and all the variables
p^2-33p+16=0
a = 1; b = -33; c = +16;
Δ = b2-4ac
Δ = -332-4·1·16
Δ = 1025
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-5\sqrt{41}}{2*1}=\frac{33-5\sqrt{41}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+5\sqrt{41}}{2*1}=\frac{33+5\sqrt{41}}{2} $
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