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4p^2-2=0
a = 4; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·4·(-2)
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*4}=\frac{0-4\sqrt{2}}{8} =-\frac{4\sqrt{2}}{8} =-\frac{\sqrt{2}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*4}=\frac{0+4\sqrt{2}}{8} =\frac{4\sqrt{2}}{8} =\frac{\sqrt{2}}{2} $
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