To find the solutions of the quadratic equation [tex]\(3x^2 + 14x + 16 = 0\)[/tex], we use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\(a = 3\)[/tex], [tex]\(b = 14\)[/tex], and [tex]\(c = 16\)[/tex].
1. Calculate the discriminant ([tex]\(\Delta\)[/tex]) which is [tex]\(b^2 - 4ac\)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = 14^2 - 4 \cdot 3 \cdot 16
= 196 - 192
= 4
\][/tex]
2. Find the square root of the discriminant:
[tex]\[
\sqrt{\Delta} = \sqrt{4} = 2
\][/tex]
3. Apply the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
= \frac{-(14) \pm 2}{2 \cdot 3}
= \frac{-14 \pm 2}{6}
\][/tex]
This gives us two solutions:
[tex]\[
x_1 = \frac{-14 + 2}{6} = \frac{-12}{6} = -2
\][/tex]
[tex]\[
x_2 = \frac{-14 - 2}{6} = \frac{-16}{6} = -\frac{8}{3}
\][/tex]
Therefore, the solutions to the quadratic equation [tex]\(3x^2 + 14x + 16 = 0\)[/tex] are:
[tex]\[
x = -2 \quad \text{and} \quad x = -\frac{8}{3}
\][/tex]
So, the correct choice is:
[tex]\[
x = -2, -\frac{8}{3}
\][/tex]
