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14p^2+8p-40=0
a = 14; b = 8; c = -40;
Δ = b2-4ac
Δ = 82-4·14·(-40)
Δ = 2304
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{2304}=48$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-48}{2*14}=\frac{-56}{28} =-2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+48}{2*14}=\frac{40}{28} =1+3/7 $
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