$a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0 $ Let $a,\,b,\,c,\,d$ be distinct real numbers and $a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0$. Then find the value of $a+b+c+d$.
I  could only get $2$ equations that 
$a=2c-b$ and $c=2a-d$.

Vieta's formulas $\Rightarrow \ \ $ $a+b=2c$, $c+d=2a$ $\ \ \Rightarrow \ \ $ $a+b+c+d=2(a+c)$ $\ \Rightarrow \ $ $a+c=b+d$.
Denote
$$m = \dfrac{a+c}{2}=\dfrac{b+d}{2}, \qquad  p=\dfrac{c-a}{2}\color{gray}{=m-a=c-m}.$$
Then
$$
\left\{
\begin{array}{r}
a = m-p, \quad b=m+3p, \\
c = m+p, \quad d=m-3p.
\end{array}
\right.
$$
Vieta's formulas $\Rightarrow \ $ $ab=-5d$, $\ \ $ $cd=-5b$ $\ \ \Rightarrow$
$$
\left\{
\begin{array}{r}
(m-p)(m+3p)=-5m+15p, \\
(m+p)(m-3p)=-5m-15p;
\end{array}
\right.
$$
$$
\left\{
\begin{array}{r}
m^2+2mp-3p^2=-5m+15p, \\
m^2-2mp-3p^2=-5m-15p;
\end{array}
\right.
$$ 
$$
\left\{
\begin{array}{c}
m^2-3p^2=-5m, \\
2mp=15p.
\end{array}
\right.
$$ 
Since $a,b,c,d$ are distinct, then $p\ne 0$, then $2m=15$, then 
$$\color{#660011}{\Large{a+b+c+d=4m=30}}.$$
Note:
$3p^2=m^2+5m=\dfrac{375}{4}$ $\ \ \Rightarrow \ \ $ $p =\pm \dfrac{5\sqrt{5}}{2}$.
