Please, see the attached document, for I was unable to write it using Latex.
The question is:
How to solve $\displaystyle 9x^2 + 6x - 8 = 0$.
The approach you're trying to use is completing the square. You've done well to get to here:
$\displaystyle 9\left[ \left(x + \frac{1}{3}\right)^2 - 1 \right] = 0$.
Unfortunately you then proceed to undo all this, effectively going in a circle and arriving back to $\displaystyle 9x^2 + 6x - 8 = 0$.
Here is what you must do. Divide both sides of $\displaystyle 9\left[ \left(x + \frac{1}{3}\right)^2 - 1 \right] = 0$ by 9:
$\displaystyle \left(x + \frac{1}{3}\right)^2 - 1 = 0$
$\displaystyle \Rightarrow \left(x + \frac{1}{3}\right)^2 - 1^2 = 0$
which is clearly a difference of two squares.
From your previous posts I know that you (should) know how to factorise this and hence get the solution.
How to solve .
A better approach...
$\displaystyle 9x^2+6x-8=0$
Multiply a 4 to LHS and RHS
It becomes $\displaystyle 36x^2+24x-32=0$
$\displaystyle \implies(6x)^2+2\cdot 6x\cdot 2+4-36=0$
$\displaystyle \implies(6x+2)^2-6^2=0$
$\displaystyle \implies (6x-4)(6x+8)=0$
$\displaystyle \implies x=\left\{\frac 2{3},\frac {-4}{3}\right\}$