**Answer:**

So by using the quadratic formula, we find the answer:

\[

x=3 \quad \text { or } \quad x=-\frac{1}{2}

\]

**Explanation:**

**Step 1: Find \(a, b, c\)**

Our equation looks like:

\[

a x^{2}+b x+c=0

\]

Let's find \(a, b, c\) :

- \(a\) is the coefficient of \(x^{\wedge} 2\)

- \(b\) is the coefficient of \(x\)

- \(\mathrm{c}\) is the constant term

So for our equation:

\[

\begin{gathered}

2 x^{2}-5 x-3=0 \\

2 x^{2}+-5 x+-3=0 \\

a=2 \\

b=-5 \\

c=-3

\end{gathered}

\]

**Step 2: Plug in \(a, b, c\)**

Let's plug in values \(a=2, b=-5, c=-3\) to solve \(2 x^{\wedge} 2-5 x-3=0\) :

\[

\begin{gathered}

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\

x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(2)(-3)}}{2(2)}

\end{gathered}

\]

**Step 3: Simplify**

We then simplify step-by-step:

\[

\begin{gathered}

x=\frac{5 \pm \sqrt{25-(-24)}}{4} \\

x=\frac{5 \pm \sqrt{49}}{4} \\

x=\frac{5 \pm 7}{4}

\end{gathered}

\]

Separate the plus-minus symbol into two solutions:

\(x=\frac{5+7}{4} \quad\) or \(\quad x=\frac{5-7}{4}\)

\(x=\frac{12}{4} \quad\) or \(\quad x=\frac{-2}{4}\)

\(x=3 \quad\) or \(\quad x=-\frac{1}{2}\)

**Answer**

So by using the quadratic formula, we find the answer:

\[

x=3 \quad \text { or } \quad x=-\frac{1}{2}

\]