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8p^2=1
We move all terms to the left:
8p^2-(1)=0
a = 8; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·8·(-1)
Δ = 32
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*8}=\frac{0-4\sqrt{2}}{16} =-\frac{4\sqrt{2}}{16} =-\frac{\sqrt{2}}{4} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*8}=\frac{0+4\sqrt{2}}{16} =\frac{4\sqrt{2}}{16} =\frac{\sqrt{2}}{4} $
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