3q^2+4q=2

Solution for 3q^2+4q=2 equation:

3q^2+4q=2

We move all terms to the left:

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

a = 3; b = 4; c = -2;
Δ = b2-4ac
Δ = 42-4·3·(-2)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*3}=\frac{-4-2\sqrt{10}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*3}=\frac{-4+2\sqrt{10}}{6}
$