Solve the inequality \(2x^{2} + 5x - 3 \geq 0\).

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 \(x \geq \frac{1}{2}\)