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p^2+60p-700=0
a = 1; b = 60; c = -700;
Δ = b2-4ac
Δ = 602-4·1·(-700)
Δ = 6400
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{6400}=80$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-80}{2*1}=\frac{-140}{2} =-70 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+80}{2*1}=\frac{20}{2} =10 $
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