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p^2+15p-100=0
a = 1; b = 15; c = -100;
Δ = b2-4ac
Δ = 152-4·1·(-100)
Δ = 625
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{625}=25$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-25}{2*1}=\frac{-40}{2} =-20 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+25}{2*1}=\frac{10}{2} =5 $
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