Answer
$g=7;\quad x = - \frac{111}{17}$
Work Step by Step
Given \begin{equation}
3.4 g^2-1.6 g-8.4=147.
\end{equation} This equation can be best solved by the quadratic formula. First bring the equation to the standard form:
\begin{equation}
\begin{aligned}
ax^2+bx+c&=0\\
3.4 g^2-1.6 g-8.4&=147 \\
\left(3.4 g^2-1.6 g-155.4\right)\cdot 10&=0\cdot 10\\
34g^2-16g-1554&= 0.
\end{aligned}
\end{equation} Apply the quadratic formula: \begin{equation}
\begin{aligned}
a & =34, b=-16, c=-1554 \\
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
g & =\frac{-(-16) \pm \sqrt{(-16)^2-4 \cdot 34(-1554)}}{2 \cdot 34}\\
& =\frac{16 \pm 460}{4\cdot 17}\\
& = \frac{4\pm115}{17}\\
\Longrightarrow g & =\frac{4+117}{17} \\
& =7 \\
g & =\frac{4+117}{17} \\
& =-\frac{111}{17} \\
& \approx -6.52941
\end{aligned}
\end{equation} Check \begin{equation}
\begin{aligned}
3.4 \cdot 7^2-1.6\cdot 7-8.4& \stackrel{?}{=} 147 \\
147 & =147 \checkmark \\
3.4 \cdot (-6.529)^2-1.6\cdot (-6.529)-8.4& \stackrel{?}{=} 147 \\
147 & =147 \checkmark.
\end{aligned}
\end{equation} The solution is \begin{equation}
\begin{aligned}
g&=7\\
g& = - \frac{111}{17}.
\end{aligned}
\end{equation}
