To solve this inequality, we can start by finding the roots of the quadratic equation \(2x^2+5x-3=0\). We can do this by factoring the quadratic or by using the quadratic formula. Factoring the quadratic, we get:
\begin{align*}
2x^2+5x-3&=0\\
(2x-1)(x+3)&=0
\end{align*}
So the roots are \(x=\frac{1}{2}\) and \(x=-3\). These are the values of \(x\) that make the left-hand side of the inequality equal to zero.
Now, we can use these roots to determine the sign of the expression \(2x^2+5x-3\) for different values of \(x\). We can do this by testing values of \(x\) in the intervals between the roots and outside the roots.
For example, if we pick a value of \(x\) less than \(-3\), say \(x=-4\), then we have:
\begin{align*}
2x^2+5x-3&=2(-4)^2+5(-4)-3\\
&=32-20-3\\
&=9
\end{align*}
Since \(9\) is positive, this means that any value of \(x\) less than \(-3\) makes the expression \(2x^2+5x-3\) positive.
Similarly, if we pick a value of \(x\) between \(-3\) and \(\frac{1}{2}\), say \(x=0\), then we have:
\begin{align*}
2x^2+5x-3&=2(0)^2+5(0)-3\\
&=-3
\end{align*}
Since \(-3\) is negative, this means that any value of \(x\) between \(-3\) and \(\frac{1}{2}\) makes the expression \(2x^2+5x-3\) negative.
Finally, if we pick a value of \(x\) greater than \(\frac{1}{2}\), say \(x=1\), then we have:
\begin{align*}
2x^2+5x-3&=2(1)^2+5(1)-3\\
&=4+5-3\\
&=6
\end{align*}
Since \(6\) is positive, this means that any value of \(x\) greater than \(\frac{1}{2}\) makes the expression \(2x^2+5x-3\) positive.
Putting all of this together, we can see that the inequality \(2x^2+5x-3\geq 0\) is true when \(x\leq -3\) or \(x\geq \frac{1}{2}\). Therefore, the correct option is:
- \(x \leq -3\) or \(x \geq \frac{1}{2}\)