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p2+13p+22=0
We add all the numbers together, and all the variables
p^2+13p+22=0
a = 1; b = 13; c = +22;
Δ = b2-4ac
Δ = 132-4·1·22
Δ = 81
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{81}=9$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-9}{2*1}=\frac{-22}{2} =-11 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+9}{2*1}=\frac{-4}{2} =-2 $
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