Answer
$x=-\dfrac{5}{3}
\text{ and }
x=\dfrac{1}{3}
$
Work Step by Step
Using the properties of equality, the given equation, $
12x+9x^2=5
,$ is equivalent to
\begin{align*}
12x+9x^2-5&=0
\\
9x^2+12x-5&=0
\end{align*}
In the equation above, $a=
9
,$ $b=
12
,$ and $c=
-5
.$
Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}\require{cancel}x&=
\dfrac{-12\pm\sqrt{12^2-4(9)(-5)}}{2(9)}
\\\\&=
\dfrac{-12\pm\sqrt{144+180}}{18}
\\\\&=
\dfrac{-12\pm\sqrt{324}}{18}
\\\\&=
\dfrac{-12\pm18}{18}
\end{align*}
\begin{array}{lcl}
&\Rightarrow
\dfrac{-12-18}{18} &\text{ OR }& \dfrac{-12+18}{18}
\\\\&
=\dfrac{-30}{18} && =\dfrac{6}{18}
\\\\&
=-\dfrac{5}{3} && =\dfrac{1}{3}
\end{array}
Hence, the solutions are $
x=-\dfrac{5}{3}
\text{ and }
x=\dfrac{1}{3}
.$
