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p2-10p+24=0
We add all the numbers together, and all the variables
p^2-10p+24=0
a = 1; b = -10; c = +24;
Δ = b2-4ac
Δ = -102-4·1·24
Δ = 4
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{4}=2$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*1}=\frac{8}{2} =4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*1}=\frac{12}{2} =6 $
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