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p2+11p+28=0
We add all the numbers together, and all the variables
p^2+11p+28=0
a = 1; b = 11; c = +28;
Δ = b2-4ac
Δ = 112-4·1·28
Δ = 9
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{9}=3$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*1}=\frac{-14}{2} =-7 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*1}=\frac{-8}{2} =-4 $
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