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 \ \ $ $p =\pm \dfrac{5\sqrt{5}}{2}$.
Best Answer
Here is a method based on how I was first taught to solve quadratics.
Note that $2b^2-ab-6a^2=(2b+3a)(b-2a)$
Then look for two expressions which have
product $abc^2(2b+3a)(b-2a)$ [coefficient of $x^2$ times the constant term]
and whose sum is $3a^2c+b^2c$ [coefficient of $x$]
The sum is homogeneous degree three and divisible by $c$, and there is no $ab$ term, so this needs to cancel. The only possibility is easily seen to be $ac(2b+3a) + bc (b-2a)=3a^2c+b^2c$
So the quadratic can be rewritten $$abc^2x^2+ac(2b+3a)x+bc(b-2a)x+(2b+3a)(b-2a)$$and this becomes $$acx \left(bcx+2b+3a\right)+(b-2a)(bcx+2b+3a)=(acx+b-2a)(bcx+2b+3a)$$
It is sometimes almost automatic to compute the discriminant and show that it is a rational square, or to use the quadratic formula. Here it is quite easy to check whether there is an easy spot for the sum/product procedure and fall back on the more general methods if you don't spot one. This method has easier manipulations than the others suggested.