what are the solutions of 3x² + 14x + 16 = 0?\no x = -8/3, -2\no x = -2, -3/8\no x = 3/8, 2\no x = 8/3, 2

Explanation:

Step1: Use quadratic - formula

For a quadratic equation $ax^{2}+bx + c = 0$ ($a\neq0$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In the equation $3x^{2}+14x + 16 = 0$, $a = 3$, $b=14$, and $c = 16$.

Step2: Calculate the discriminant $\Delta=b^{2}-4ac$

$\Delta=(14)^{2}-4\times3\times16=196 - 192=4$.

Step3: Find the roots

$x=\frac{-14\pm\sqrt{4}}{2\times3}=\frac{-14\pm2}{6}$. When we take the plus - sign: $x=\frac{-14 + 2}{6}=\frac{-12}{6}=-2$. When we take the minus - sign: $x=\frac{-14-2}{6}=\frac{-16}{6}=-\frac{8}{3}$.

Answer:

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