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35p^2+4p-28p-4=0
We add all the numbers together, and all the variables
35p^2-24p-4=0
a = 35; b = -24; c = -4;
Δ = b2-4ac
Δ = -242-4·35·(-4)
Δ = 1136
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{71}}{2*35}=\frac{24-4\sqrt{71}}{70} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{71}}{2*35}=\frac{24+4\sqrt{71}}{70} $
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