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7p^2-14p-56=0
a = 7; b = -14; c = -56;
Δ = b2-4ac
Δ = -142-4·7·(-56)
Δ = 1764
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{1764}=42$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-42}{2*7}=\frac{-28}{14} =-2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+42}{2*7}=\frac{56}{14} =4 $
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