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p^2=13p-36
We move all terms to the left:
p^2-(13p-36)=0
We get rid of parentheses
p^2-13p+36=0
a = 1; b = -13; c = +36;
Δ = b2-4ac
Δ = -132-4·1·36
Δ = 25
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{25}=5$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*1}=\frac{8}{2} =4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*1}=\frac{18}{2} =9 $
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