Answer
The solution set is $\left\{\frac{3}{2}\right\}$.
Work Step by Step
Subtract $12x$ to both sides of the equation to obtain:
$4x^2-12x+9=0$
RECALL:
A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$
If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping.
The trinomial in the equation has $a=4$, $b=-12$, and $c=9$.
Thus, $ac = 4(9) = 36$
Note that $36=-6(-6)$ and $-12 = (-6)+(-6)$
This means that $d=-6$ and $e=-6$.
Rewrite the middle term of the trinomial as $-6x$ +$(-6x)$ to obtain:
$4x^2-12x+9 = 0
\\4x^2+(-6x)+(-6x)+9 = 0
\\4x^2-6x+(-6x) +9=0$
Regroup then factor out the GCF in each group to obtain:
$(4x^2-6x)+(-6x+9)=0
\\2x(2x-3) +(-3)(2x-3)=0$
Factor out $2x-3$ to obtain:
$(2x-3)(2x-3)=0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
$\begin{array}{ccc}
&2x-3=0 &\text{ or } &2x-3=0
\\&2x=3 &\text{ or } &2x=3
\\&x=\frac{3}{2} &\text{ or } &x=\frac{3}{2}
\end{array}$
Therefore the solution set is $\left\{\frac{3}{2}\right\}$.
