We are asked to solve the quadratic equation $2x^2 - 5x - 3 = 0$.

1. Problem Description

We are asked to solve the quadratic equation $2x^2 - 5x - 3 = 0$.

2. Solution Steps

We can solve this quadratic equation using the quadratic formula or by factoring. Let's try factoring.

We are looking for two numbers that multiply to $(2)(-3) = -6$ and add up to $-5$. Those two numbers are $-6$ and $1$. We can rewrite the middle term as $-6x + x$.

$2x^2 - 5x - 3 = 2x^2 - 6x + x - 3$

Now, we can factor by grouping.

$2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)$

Setting this equal to zero gives us:

$(2x + 1)(x - 3) = 0$

So, either $2x + 1 = 0$ or $x - 3 = 0$.

If $2x + 1 = 0$, then $2x = -1$, so $x = -\frac{1}{2}$.

If $x - 3 = 0$, then $x = 3$.

Alternatively, we can use the quadratic formula:

For a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions are given by:

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

In our case, $a = 2$, $b = -5$, and $c = -3$.

Plugging these values into the quadratic formula gives:

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

$x = \frac{5 \pm \sqrt{25 + 24}}{4}$

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

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

Thus, the two possible values for $x$ are:

$x = \frac{5 + 7}{4} = \frac{12}{4} = 3$

$x = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}$

3. Final Answer

The solutions are $x = 3$ and $x = -\frac{1}{2}$.