Evaluate $x$ for $2x^2 - 3x - 2<0$ Solve $$2x^2-3x-2=0$$ we get $x=-0.5;2$

So, $$-0.5&lt;x&lt;2$$
The key answer is $$0.5&lt;x&lt;2$$
Which one is true?
Factorizing, we get $$2x^2-3x-2=(2x+1)(x-2).$$

Its roots are $-\frac12$ and $2$.  Using a tabular method, we have
\begin{array}{|c|c|c|c|}
\hline
\text{Interval} &amp; 2x+1 &amp; x-2 &amp; \text{Sign of}\ f(x)\\
\hline
x&lt;-\frac12 &amp; - &amp; - &amp; +\\
\hline
x=-\frac12 &amp; 0 &amp; - &amp; 0\\
\hline
\color{blue}{-\frac12 &lt; x &lt; 2} &amp; + &amp; - &amp; \color{blue}{-}\\
\hline
x=2 &amp; + &amp; 0 &amp; 0\\
\hline
x&gt;2 &amp; + &amp; + &amp; +\\
\hline
\end{array}

Therefore, the answer should be $\boxed{-\frac12&lt;x&lt;2}$.
The correct answer is $-0.5&lt;x&lt;2$.  There should be a negative sign in the key answer.

The inequality holds for all $x$ between the zeroes, including for example $x=0$.