Answer
$x=-\dfrac{3}{4}$ OR $x=1$
Work Step by Step
In the form $x^2+bx=c,$ the given equation, $
4x^2-x-3=0
,$ is equivalent to
\begin{align*}
4x^2-x-3+3&=0+3
\\
4x^2-x&=3
\\\\
\dfrac{4x^2-x}{4}&=\dfrac{3}{4}
\\\\
\dfrac{4}{4}x^2-\dfrac{1}{4}x&=\dfrac{3}{4}
\\\\
x^2-\dfrac{1}{4}x&=\dfrac{3}{4}
.\end{align*}
Adding $\left( \dfrac{b}{2} \right)^2$ on both sides to complete the square of the left side results to
\begin{align*}
x^2-\dfrac{1}{4}x+\left(-\dfrac{1/4}{2} \right)^2&=\dfrac{3}{4}+\left(-\dfrac{1/4}{2} \right)^2
\\\\
x^2-\dfrac{1}{4}x+\left(-\dfrac{1}{8} \right)^2&=\dfrac{3}{4}+\left(-\dfrac{1}{8} \right)^2
\\\\
x^2-\dfrac{1}{4}x+\dfrac{1}{64}&=\dfrac{3}{4}+\dfrac{1}{64}
\\\\
\left(x-\dfrac{1}{8}\right)^2&=\dfrac{48}{64}+\dfrac{1}{64}
\\\\
\left(x-\dfrac{1}{8}\right)^2&=\dfrac{49}{64}
.\end{align*}
Taking the square root of both sides (Square Root Property) and then solving for the variable, the equation above is equivalent to
\begin{align*}
x-\dfrac{1}{8}&=\pm\sqrt{\dfrac{49}{64}}
\\\\
x-\dfrac{1}{8}&=\pm\dfrac{7}{8}
\\\\
x&=\dfrac{1}{8}\pm\dfrac{7}{8}
\end{align*}
\begin{array}{lcl}
x=\dfrac{1}{8}-\dfrac{7}{8} &\text{ OR }& x=\dfrac{1}{8}+\dfrac{7}{8}
\\\\
x=-\dfrac{6}{8} && x=\dfrac{8}{8}
\\\\
x=-\dfrac{3}{4} && x=1
.\end{array}
Hence, the solutions are
$x=-\dfrac{3}{4}$ OR $x=1$.