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p2-9p+6=0
We add all the numbers together, and all the variables
p^2-9p+6=0
a = 1; b = -9; c = +6;
Δ = b2-4ac
Δ = -92-4·1·6
Δ = 57
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{-(-9)-\sqrt{57}}{2*1}=\frac{9-\sqrt{57}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{57}}{2*1}=\frac{9+\sqrt{57}}{2} $
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