8r^2-3=0

Solution for 8r^2-3=0 equation:

8r^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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$r_{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}
$
$r_{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}
$