Answer
The solution set is $\left\{\frac{1}{2}, \frac{3}{2}\right\}$.
Work Step by Step
Distribute $2$ to obtain:
$4u^2-8u+3=0$
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 $a=4$, $b=-8$, and $c=3$.
Thus, $ac = 4(3) = 12$
Note that $12=-6(-2)$ and $-8 = (-6)+(-2)$
This means that $d=-6$ and $e=-2$.
Rewrite the middle term of the trinomial as $-6u$ +$(-2u)$ to obtain:
$4u^2-8u+3 = 0
\\4u^2+(-6u)+(-2u)+3 = 0
\\4u^2-6u+(-2u) +3=0$
Regroup then factor out the GCF in each group to obtain:
$(4u^2-6u)+(-2u+3)=0
\\2u(2u-3) +(-1)(2u-3)=0$
Factor out $2u-3$ to obtain:
$(2u-3)(2u-1)=0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
$\begin{array}{ccc}
&2u-3=0 &\text{ or } &2u-1=0
\\&2u=3 &\text{ or } &2u=1
\\&u=\frac{3}{2} &\text{ or } &u=\frac{1}{2}
\end{array}$
Therefore the solution set is $\left\{\frac{1}{2}, \frac{3}{2}\right\}$.
