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p^2+10p-56=0
a = 1; b = 10; c = -56;
Δ = b2-4ac
Δ = 102-4·1·(-56)
Δ = 324
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{324}=18$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-18}{2*1}=\frac{-28}{2} =-14 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+18}{2*1}=\frac{8}{2} =4 $
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