How do you solve $2{x^2} - 3x - 9 = 0$?

Hint:There are various methods by which we can solve the given equation i.e., factorization, completing the square or quadratic formula. Let us solve the given equation for the value of $x$ by quadratic formula. Before solving this question, we will first have to compare the given equation with the standard quadratic equation, which is $a{x^2} + bx + c = 0$, where$a \ne 0$. After comparing both the equations with each other, we will have to find the values of $a,b$ and $c$. Next, we will have to substitute the obtained values in the quadratic formula.
Formula used:
Quadratic formula: $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
The quadratic equation is $2{x^2} - 3x - 9 = 0$.
We have to solve the given equation for finding out the value of $x$. Before starting with the solution, we first need to compare the given equation with the standard quadratic equation.
Standard quadratic equation: $a{x^2} + bx + c = 0$.
Now, let $2{x^2} - 3x - 9 = 0$-----(1)
After comparing equation (1) with the standard quadratic equation we get, $a = 2,b = - 3$ and $c = - 9$
As for the next step, we have to substitute the obtained values of $a,b$ and $c$ in the quadratic formula in order to find the values of $x$.
After substituting the values, we get:
$
\Rightarrow \dfrac{{ - \left( { - 3} \right) \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\left( 2 \right)\left(
{ - 9} \right)} }}{{2\left( 2 \right)}} \\
\Rightarrow \dfrac{{3 \pm \sqrt {9 + 72} }}{4} \\
\Rightarrow \dfrac{{3 \pm \sqrt {81} }}{4} \\
\Rightarrow \dfrac{{3 \pm 9}}{4} \\
\therefore x = \dfrac{{3 + 9}}{4},\dfrac{{3 - 9}}{4} \\
x = \dfrac{{12}}{4}, - \dfrac{6}{4} \\
x = 3, - \dfrac{3}{2} \\
$
Thus, we can say that $x = 3$ or $x = - \dfrac{3}{2}$.
Hence, the quadratic equation $2{x^2} - 3x - 9 = 0$ after solving with quadratic formula has roots $x = 3$or$x = - \dfrac{3}{2}$.

Note: In the quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, the expression $\sqrt {{b^2} - 4ac} $ is called as discriminant and is often denoted by $D$. Now, depending upon the quadratic equation, the value of discriminant changes and therefore, the value of roots also change. If $D$ is positive or greater than zero, then the two roots of the equation are real. If $D$ is zero, then roots are real but if $D$ is negative or less than zero, then roots are not real.