3q(+4/3)=2

Solution for 3q(+4/3)=2 equation:

3q(+4/3)=2

We move all terms to the left:

3q(+4/3)-(2)=0

We multiply parentheses

12q^2-2=0

a = 12; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·12·(-2)
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*12}=\frac{0-4\sqrt{6}}{24}
=-\frac{4\sqrt{6}}{24}
=-\frac{\sqrt{6}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*12}=\frac{0+4\sqrt{6}}{24}
=\frac{4\sqrt{6}}{24}
=\frac{\sqrt{6}}{6}
$