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2p^2-14p-16=0
a = 2; b = -14; c = -16;
Δ = b2-4ac
Δ = -142-4·2·(-16)
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$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-18}{2*2}=\frac{-4}{4} =-1 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+18}{2*2}=\frac{32}{4} =8 $
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