Algebra help

There are two pairs $(x,y)$ of real numbers that satisfy the equation $x + y = 4xy = 5$.
Given that the solutions x are in the form $x = \dfrac{a \pm b \sqrt{c}}{d}$ ,
where a, b, c, and d are positive integers and the expression is completely simplified,
what is the value of $a + b + c + d$?

My attempt:

$\text{Let $a=x+y=5$ } \\ \text{Let $b=xy=\dfrac{5}{4}$ }$

$\begin{array}{|rcll|} \hline x^2 - Ax + B &=& 0 \\ x^2 - 5x + \dfrac{5}{4} &=& 0 \\\\ x &=& \dfrac{ 5 \pm \sqrt{25-4\dfrac{5}{4}} }{2} \\\\ x &=& \dfrac{5 \pm \sqrt{25-5} }{2} \\\ x &=& \dfrac{5 \pm \sqrt{20} }{2} \\\\ \mathbf{x} &=& \mathbf{\dfrac{5 \pm 2\sqrt{5} }{2}} \\ \hline x+y &=& 5 \\ y &=& 5-x \\ y &=& 5-\dfrac{5 \pm 2\sqrt{5} }{2} \\\ y &=& \dfrac{10-5 \mp 2\sqrt{5} }{2} \\\\ \mathbf{y} &=& \mathbf{\dfrac{5 \mp 2\sqrt{5} }{2}} \\ \hline \end{array}$

Two pairs (x,y):
$\begin{array}{|rcll|} \hline (x_1,~y_1) &=& \left(\dfrac{5 + 2\sqrt{5} }{2},~\dfrac{5 - 2\sqrt{5} }{2} \right) \\\\ (x_2,~y_2) &=& \left(\dfrac{5 - 2\sqrt{5} }{2},~\dfrac{5 + 2\sqrt{5} }{2} \right) \\ \hline \end{array}$

$a=5,~ b=2,~ c=5,~ d=2 \\ \mathbf{a+b+c+d = 14}$