Question

what are the solutions of $3x^2 + 14x + 16 = 0?LXB0\bigcirc$ $x = -2, -\frac{3}{8}LXB1\bigcirc$ $x = \frac{8}{3}, 2$

Explanation:

Step1: Identify quadratic coefficients

For $3x^2 + 14x + 16 = 0$, we have $a=3$, $b=14$, $c=16$.

Step2: Apply quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute values: $x=\frac{-14\pm\sqrt{14^2-4(3)(16)}}{2(3)}$

Step3: Calculate discriminant

$\sqrt{196 - 192}=\sqrt{4}=2$

Step4: Compute two solutions

First solution: $x=\frac{-14+2}{6}=\frac{-12}{6}=-2$
Second solution: $x=\frac{-14-2}{6}=\frac{-16}{6}=-\frac{8}{3}$

Answer:

$x = -\frac{8}{3}, -2$