Answer
$-\frac{3}{4}\text{ or }0$.
Work Step by Step
$8{{x}^{2}}+6x=0$
Take the common multiplier as,
$\begin{align}
& 8{{x}^{2}}+6x=0 \\
& 2x\left( 4x+3 \right)=0
\end{align}$
Thus, $2x=0$
Or,
$\left( 4x+3 \right)=0$
Now rewrite the term $2x=0$ and further simplify,
$2x=0$
Now divide by $2$ on both sides:
$\begin{align}
& 2x=0 \\
& \frac{2x}{2}=\frac{0}{2} \\
& x=0
\end{align}$
Now, rewrite the term $\left( 4x+3 \right)=0$ and further simplify:
$\left( 4x+3 \right)=0$
Subtract by $3$ on both sides:
$\begin{align}
& 4x+3-3=-3 \\
& 4x=-3
\end{align}$
Now, divide by $4$ on both sides:
$\begin{align}
& 4x=-3 \\
& \frac{4x}{4}=-\frac{3}{4} \\
& x=-\frac{3}{4}
\end{align}$
Thus, the value of $x$ from the provided expression is $-\frac{3}{4}\text{ or }0$.
