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p.2(p+1)=16
We move all terms to the left:
p.2(p+1)-(16)=0
We multiply parentheses
p^2+p-16=0
a = 1; b = 1; c = -16;
Δ = b2-4ac
Δ = 12-4·1·(-16)
Δ = 65
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{65}}{2*1}=\frac{-1-\sqrt{65}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*1}=\frac{-1+\sqrt{65}}{2} $
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