3r(3r-4)=2

Solution for 3r(3r-4)=2 equation:

3r(3r-4)=2

We move all terms to the left:

3r(3r-4)-(2)=0

We multiply parentheses

9r^2-12r-2=0

a = 9; b = -12; c = -2;
Δ = b2-4ac
Δ = -122-4·9·(-2)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{6}}{2*9}=\frac{12-6\sqrt{6}}{18}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{6}}{2*9}=\frac{12+6\sqrt{6}}{18}
$