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p^2-20p-125=0
a = 1; b = -20; c = -125;
Δ = b2-4ac
Δ = -202-4·1·(-125)
Δ = 900
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{900}=30$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-30}{2*1}=\frac{-10}{2} =-5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+30}{2*1}=\frac{50}{2} =25 $
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