-8j^2+6=0

Solution for -8j^2+6=0 equation:

-8j^2+6=0

a = -8; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-8)·6
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-8}=\frac{0-8\sqrt{3}}{-16}
=-\frac{8\sqrt{3}}{-16}
=-\frac{\sqrt{3}}{-2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-8}=\frac{0+8\sqrt{3}}{-16}
=\frac{8\sqrt{3}}{-16}
=\frac{\sqrt{3}}{-2}
$