Answer
$x=\left\{ \dfrac{1-2\sqrt{5}}{2},\dfrac{1+2\sqrt{5}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
4r^2-4r-19=0
,$ express in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
The quadratic equation above has $a=
4
, b=
-4
, c=
-19
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(4)(-19)}}{2(4)}
\\\\
x=\dfrac{4\pm\sqrt{16+304}}{8}
\\\\
x=\dfrac{4\pm\sqrt{320}}{8}
.\end{array}
Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to
\begin{array}{l}\require{cancel}
x=\dfrac{4\pm\sqrt{64\cdot5}}{8}
\\\\
x=\dfrac{4\pm\sqrt{(8)^2\cdot5}}{8}
\\\\
x=\dfrac{4\pm8\sqrt{5}}{8}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel4(1)\pm\cancel4(2)\sqrt{5}}{\cancel4(2)}
\\\\
x=\dfrac{1\pm2\sqrt{5}}{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{1-2\sqrt{5}}{2}
\\\\\text{OR}\\\\
x=\dfrac{1+2\sqrt{5}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{1-2\sqrt{5}}{2},\dfrac{1+2\sqrt{5}}{2} \right\}
.$
