Answer
The solution set is\[\left\{ -\frac{1}{3},\frac{5}{3} \right\}\].
Work Step by Step
The given equation can be written as \[9{{x}^{2}}-12x-5=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=9,\,\,b=-12,\text{ and }c=-5\].
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{12\pm \sqrt{{{\left( -12 \right)}^{2}}-4\times 9\times \left( -5 \right)}}{2\times 9} \\
& =\frac{12\pm \sqrt{144+180}}{18} \\
& =\frac{12\pm \sqrt{324}}{18} \\
& =\frac{12\pm 18}{18}
\end{align}\]
Further simplify it to get,
\[\begin{align}
& x=\frac{12\pm 18}{18} \\
& =\frac{30}{18},-\frac{6}{18} \\
& =\frac{5}{3},-\frac{1}{3}
\end{align}\]
Hence, the solution set is\[\left\{ -\frac{1}{3},\frac{5}{3} \right\}\].
