Answer
the solution set is\[\left\{ -\frac{2}{3},\frac{3}{2} \right\}\].
Work Step by Step
The given equation can be written as \[6{{x}^{2}}-5x-6=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=6,\,\,b=-5,\text{ and }c=-6\].
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{5\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 6\times \left( -6 \right)}}{2\times 6} \\
& =\frac{5\pm \sqrt{25+144}}{12} \\
& =\frac{5\pm \sqrt{169}}{12} \\
& =\frac{5\pm 13}{12}
\end{align}\]
Further simplify it to get,
\[\begin{align}
& x=\frac{5\pm 13}{12} \\
& =\frac{5+13}{12},\frac{5-13}{12} \\
& =\frac{18}{12},-\frac{8}{12} \\
& =\frac{3}{2},-\frac{2}{3}
\end{align}\]
Hence, the solution set is\[\left\{ -\frac{2}{3},\frac{3}{2} \right\}\].
