-6q2+4=0

Solution for -6q2+4=0 equation:

-6q^2+4=0

a = -6; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-6)·4
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-6}=\frac{0-4\sqrt{6}}{-12}
=-\frac{4\sqrt{6}}{-12}
=-\frac{\sqrt{6}}{-3}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-6}=\frac{0+4\sqrt{6}}{-12}
=\frac{4\sqrt{6}}{-12}
=\frac{\sqrt{6}}{-3}
$