Answer
$\frac{-9+\sqrt{489}}{6}$; $\frac{-9-\sqrt{489}}{6}$
Work Step by Step
Given the equation $$3 x^2+9 x-20=14.$$ Rearrange the equation before applying the quadratic formula, then determine the constants $a$, $b$, $c$. \begin{equation}
\begin{aligned}
3 x^2+9 x-20 &=14\\
3 x^2+9 x-20-14&=0\\
3 x^2+9 x-34&=0\\
a^2x+bx+c&=0\\
a =3\quad , b= 9\ , \quad c&= -34.
\end{aligned}
\end{equation} The quadratic formula gives: \begin{equation}
\begin{aligned}
x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
x &=\frac{-9 \pm \sqrt{9^2-4 \cdot 3(-34)}}{2 \cdot 3}\\
&=\frac{-9 \pm \sqrt{489}}{6}.
\end{aligned}
\end{equation} The solution is: \begin{equation}
x=\frac{-9+\sqrt{489}}{6},\quad x=\frac{-9-\sqrt{489}}{6}.
\end{equation}
