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8p^2-3=0
a = 8; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·8·(-3)
Δ = 96
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*8}=\frac{0-4\sqrt{6}}{16} =-\frac{4\sqrt{6}}{16} =-\frac{\sqrt{6}}{4} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*8}=\frac{0+4\sqrt{6}}{16} =\frac{4\sqrt{6}}{16} =\frac{\sqrt{6}}{4} $
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