[tex]4v^2+12v+9=0[/tex]
on this point we can use the formula to factor
[tex]v=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
where a is 4, b is 12 and c is 9
replacing
[tex]v=\frac{-(12)\pm\sqrt[]{(12)^2-4(4)(9)}}{2(4)}[/tex]
and solving
[tex]\begin{gathered} v=\frac{-12\pm\sqrt[]{144-144}}{8} \\ \\ v=\frac{-12\pm\sqrt[]{0}}{8} \\ \\ v=-\frac{12}{8}=-\frac{3}{2} \end{gathered}[/tex]
The root is
[tex]-\frac{3}{2}[/tex]
then the polynomial has one root -3/2 if we rewrite the polynomial is
[tex](v+\frac{3}{2})^2=0[/tex]
or
[tex](2v+3)^2=0[/tex]
