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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$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*1}=\frac{8}{2} =4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*1}=\frac{14}{2} =7 $
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