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p2+15p+50=0
We add all the numbers together, and all the variables
p^2+15p+50=0
a = 1; b = 15; c = +50;
Δ = b2-4ac
Δ = 152-4·1·50
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$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*1}=\frac{-20}{2} =-10 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*1}=\frac{-10}{2} =-5 $
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