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3p^2-14p-16=0
a = 3; b = -14; c = -16;
Δ = b2-4ac
Δ = -142-4·3·(-16)
Δ = 388
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{97}}{2*3}=\frac{14-2\sqrt{97}}{6} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{97}}{2*3}=\frac{14+2\sqrt{97}}{6} $
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