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(p2-5p+3)=0
We add all the numbers together, and all the variables
(+p^2-5p+3)=0
We get rid of parentheses
p^2-5p+3=0
a = 1; b = -5; c = +3;
Δ = b2-4ac
Δ = -52-4·1·3
Δ = 13
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{-(-5)-\sqrt{13}}{2*1}=\frac{5-\sqrt{13}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{13}}{2*1}=\frac{5+\sqrt{13}}{2} $
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