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**Hint:**There are various ways which one might use to solve an equation, but first one needs to identify the highest power of the variable to be found in the equation. In this equation, we cannot directly find the highest power of x. Multiply the expressions to get a single line expression on the LHS. Proceed to the solution accordingly.

**Complete step-by-step solution:**

In the question we can see that there is no clarity on the highest power of the variable, x.

Hence, we use the BODMAS rule to open the brackets and multiply the contents with each other.

The LHS changes and the equation becomes,

$

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

\Rightarrow 2x \times x - 1 \times x + 3 \times 2x - 1 \times 3 = 0 \\

\Rightarrow 2{x^2} - x + 6x - 3 = 0 \\

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

$

Thus, we see that the highest power of x is 2, which makes the equation a quadratic equation.

Now, we know that $x \times y = 0$ means that x and y both individually equate to zero.

So, from the question,

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

$ \Rightarrow 2x = 1,x = - 3$

$ \Rightarrow x = \dfrac{1}{2},x = - 3$

**Note:**Mid-term factorization or completing square methods often become cumbersome and are time taking to solve. Whenever dealing with fractional or imaginary values, use of the quadratic formula to solve quadratic equations is recommended. The quadratic formula is $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where a is the coefficient of ${x^2}$, b is the coefficient of x and c is the constant from the standard quadratic equation, $a{x^2} + bx + c = 0$.

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