To find the roots of the quadratic equation [tex]\(3x^2 - 14x - 5 = 0\)[/tex], we use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, the coefficients are [tex]\(a = 3\)[/tex], [tex]\(b = -14\)[/tex], and [tex]\(c = -5\)[/tex]. Substituting these into the quadratic formula:
[tex]\[
x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3}
\][/tex]
Simplify the terms inside the square root:
[tex]\[
x = \frac{14 \pm \sqrt{196 + 60}}{6}
\][/tex]
[tex]\[
x = \frac{14 \pm \sqrt{256}}{6}
\][/tex]
Since [tex]\(\sqrt{256} = 16\)[/tex]:
[tex]\[
x = \frac{14 \pm 16}{6}
\][/tex]
This gives us two potential roots:
[tex]\[
x = \frac{14 + 16}{6} = \frac{30}{6} = 5
\][/tex]
[tex]\[
x = \frac{14 - 16}{6} = \frac{-2}{6} = -\frac{1}{3}
\][/tex]
The roots are [tex]\(5\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex].
Among these, the smaller root is [tex]\(-\frac{1}{3}\)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{-\frac{1}{3}}
\][/tex]