Answer
$x=-1$ OR $x=\dfrac{1}{3}$
Work Step by Step
In the form $x^2+bx=c,$ the given equation, $
9x^2+6x+1=4
,$ is equivalent to
\begin{align*}
9x^2+6x+1-1&=4-1
\\
9x^2+6x&=3
\\\\
\dfrac{9x^2+6x}{9}&=\dfrac{3}{9}
\\\\
\dfrac{9}{9}x^2+\dfrac{6}{9}x&=\dfrac{3}{9}
\\\\
x^2+\dfrac{2}{3}x&=\dfrac{1}{3}
.\end{align*}
Adding $\left( \dfrac{b}{2} \right)^2$ on both sides to complete the square of the left side results to
\begin{align*}
x^2+\dfrac{2}{3}x+\left(\dfrac{2/3}{2} \right)^2&=\dfrac{1}{3}+\left(\dfrac{2/3}{2} \right)^2
\\\\
x^2+\dfrac{2}{3}x+\left(\dfrac{1}{3} \right)^2&=\dfrac{1}{3}+\left(\dfrac{1}{3} \right)^2
\\\\
x^2+\dfrac{2}{3}x+\dfrac{1}{9}&=\dfrac{1}{3}+\dfrac{1}{9}
\\\\
\left( x+\dfrac{1}{3} \right)^2&=\dfrac{3}{9}+\dfrac{1}{9}
\\\\
\left( x+\dfrac{1}{3} \right)^2&=\dfrac{4}{9}
.\end{align*}
Taking the square root of both sides (Square Root Property) and then solving for the variable, the equation above is equivalent to
\begin{align*}
x+\dfrac{1}{3}&=\pm\sqrt{\dfrac{4}{9}}
\\\\
x+\dfrac{1}{3}&=\pm\dfrac{2}{3}
\\\\
x&=-\dfrac{1}{3}\pm\dfrac{2}{3}
\end{align*}
\begin{array}{lcl}
x=-\dfrac{1}{3}-\dfrac{2}{3} &\text{ OR }& x=-\dfrac{1}{3}+\dfrac{2}{3}
\\\\
x=-\dfrac{3}{3} &\text{ OR }& x=\dfrac{1}{3}
\\\\
x=-1 &\text{ OR }& x=\dfrac{1}{3}
.\end{array}
Hence, the solutions are
$x=-1$ OR $x=\dfrac{1}{3}$.
