Answer
$x=-\dfrac{1}{2}
\text{ and }
x=\dfrac{3}{2}$
Work Step by Step
In the given equation,
\begin{align*}
4x^2-4x-3=0
,\end{align*} $a=
4
,$ $b=
-4
,$ and $c=
-3
.$
Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}\require{cancel}x&=
\dfrac{-(-4)\pm\sqrt{(-4)^2-4(4)(-3)}}{2(4)}
\\\\&=
\dfrac{4\pm\sqrt{16+48}}{8}
\\\\&=
\dfrac{4\pm\sqrt{64}}{8}
\\\\&=
\dfrac{4\pm8}{8}
\end{align*}
\begin{array}{lcl}
&\Rightarrow
\dfrac{4-8}{8} &\text{ OR }& \dfrac{4+8}{8}
\\\\&
=\dfrac{-4}{8} && =\dfrac{12}{8}
\\\\&
=-\dfrac{1}{2} && =\dfrac{3}{2}
\end{array}
Hence, the solutions are $
x=-\dfrac{1}{2}
\text{ and }
x=\dfrac{3}{2}
.$
