#### Answer

The required value is\[\left( 2,\ -5 \right)\]

#### Work Step by Step

The quadratic function of the form, $f\left( x \right)=a{{x}^{2}}+bx+c$ , can be converted into its standard form as $f\left( x \right)=a{{\left( x+\frac{b}{2a} \right)}^{2}}+c-\frac{{{b}^{2}}}{4a}$
The quadratic function in its standard form can be written as
$f\left( x \right)=a{{\left( x-h \right)}^{2}}+k,\,\,\,\,\,\,\,\,a\ne 0$
Here, $a,\ h,\ k$ are constants and $x$ is a variable.
The graph of $f\left( x \right)$ is a parabola which is symmetric about the line $x=h$.
Now, the coordinates of the vertex of the parabola are $\left( h,k \right)$.
Compare this with standard form to get $h=-\frac{b}{2a}$ and $k=c-\frac{{{b}^{2}}}{4a}$.
The coordinates of the vertex of the parabola can also be written as $\left( -\frac{b}{2a},f\left( -\frac{b}{2a} \right) \right)$.
Let us consider the given function $f\left( x \right)=2{{x}^{2}}-8x+3$.
Compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$ to get $a=2,\ b=-8$.
The $x$ -coordinates of vertex are
$\begin{align}
& h=-\frac{b}{2a} \\
& =-\left( \frac{-8}{4} \right) \\
& =2
\end{align}$
Calculate y-coordinate by determining $f\left( 2 \right)$:
$\begin{align}
& f\left( 2 \right)=2{{\left( 2 \right)}^{2}}-8\left( 2 \right)+3 \\
& =-5
\end{align}$
Hence, the coordinates of the vertex are $\left( 2,\ -5 \right)$