Answer
$-\frac{1}{4}\text{ and 1}$.
Work Step by Step
$f\left( x \right)=4{{x}^{2}}-3x-1$
Now, equate the expression to $f\left( x \right)=0$ as,
$4{{x}^{2}}-3x-1=0$
Now compare the equation $4{{x}^{2}}-3x-1=0$ with the standard form of the quadratic equation $a{{x}^{2}}+bx+c=0$,
Therefore,
$a=4,b=-3\text{ and }c=-1$
Substitute $a=4,b=-3\text{ and }c=-1$ into the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$,
$\begin{align}
& x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& =\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\cdot 4\cdot \left( -1 \right)}}{2\cdot 4} \\
& =\frac{3\pm \sqrt{9+16}}{8} \\
& =\frac{3\pm \sqrt{25}}{8}
\end{align}$
Simplify further as shown below,
$\begin{align}
& x=\frac{3\pm \sqrt{25}}{8} \\
& =\frac{3\pm 5}{8}
\end{align}$
Thus, $x=\frac{3+5}{8}$
And,
$x=\frac{3-5}{8}$
Now, consider the value $x=\frac{3+5}{8}$ and solve further,
$\begin{align}
& x=\frac{3+5}{8} \\
& =\frac{8}{8} \\
& =1
\end{align}$
Now, consider the value $x=\frac{3-5}{8}$ and solve further,
$\begin{align}
& x=\frac{3-5}{8} \\
& =\frac{-2}{8} \\
& =-\frac{1}{4}
\end{align}$
