How do you solve $2{{x}^{2}}-3x+4=0$ ?

Hint: In this question, we have to find the value of x. The equation given to us is in the form of a quadratic, therefore when we solve this problem, we will get two values for x, which satisfy the equation. Therefore, we will apply the discriminant method to solve this problem. We will first compare the given equation with the general form of the quadratic equation and then get the value of a, b, and c. Then, we will get the value of discriminant $D=\sqrt{{{b}^{2}}-4ac}$, and thus find the value of x using the discriminant formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .After necessary calculations, get two equations of x, we solve them separately to get the value of x, which is our required answer.

Complete step by step solution:
According to the question, a quadratic equation is given to us and we have to solve the equation for the value of x.
The equation is $2{{x}^{2}}-3x+4=0$ ----------------- (1)
As we know, the general quadratic equation is in form of $a{{x}^{2}}+bx+c=0$ ---------- (2)
Thus, on comparing equation (1) and (2), we get $a=2,$ $b=-3,$ and $c=4$ ------- (3)
So, now we will apply the discriminant formula $D=\sqrt{{{b}^{2}}-4ac}$ by putting the above values in the formula, we get
$\begin{align}
  & \Rightarrow D=\sqrt{{{(-3)}^{2}}-4.(2).(4)} \\
 & \Rightarrow D=\sqrt{9-32} \\
\end{align}$
Thus, on further solving, we get
$\Rightarrow D=\sqrt{-23}$
We see that the square root has a negative term which implies the discriminant has no real roots, thus we will remove the negative sign by putting iota $''i''$ in the above equation, we get
$\Rightarrow D=i\sqrt{23}$ -------------- (4)
Since we see the discriminant is a real number, thus now we will find the value of x, using the formula,
$\Rightarrow x=\dfrac{-b\pm D}{2a}$
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ --------------- (5)
So, we will put the value of equation (3) and (4) in equation (5), we get
$\Rightarrow x=\dfrac{-(-3)\pm i\sqrt{23}}{2.(2)}$
On further simplification, we get
$\Rightarrow x=\dfrac{3\pm i\sqrt{23}}{4}$
Therefore, we will split the above equation in terms of (+) and (-), we get
$\Rightarrow x=\dfrac{3+i\sqrt{23}}{4}$ -------- (6) , or
 $\Rightarrow x=\dfrac{3-i\sqrt{23}}{4}$ ---------- (7)
Now, we will first solve equation (6) that is we will split the denominator with respect to addition, we get
$\Rightarrow x=\dfrac{3}{4}+\dfrac{i\sqrt{23}}{4}$
Now, we will first solve equation (7) that is we will split the denominator with respect to subtraction, we get
$\Rightarrow x=\dfrac{3}{4}-\dfrac{i\sqrt{23}}{4}$
Therefore, for the equation $2{{x}^{2}}-3x+4$ , we get the value of $x=\dfrac{3}{4}+\dfrac{i\sqrt{23}}{4},\dfrac{3}{4}-\dfrac{i\sqrt{23}}{4}$ .

Note: While solving this problem, do all the steps carefully and avoid errors to get the correct answer. Do not forget, how to remove the negative sign in the square root, that is always put the iota sign while removing the negative sign.