Sum of real roots!

$\begin{aligned} x^2+7x-5 & = 5\sqrt{x^3-1} \\ x^2 + x + 1 + 6(x-1) & = 5\sqrt{(x-1)(x^2+x+1)} \\ (x^2 + x + 1) - 5\sqrt{(x-1)(x^2+x+1)} + 6(x-1) & = 0 \\ \left(\sqrt{x^2+x+1} - 2\sqrt{x-1} \right)\left(\sqrt{x^2+x+1} - 3\sqrt{x-1} \right) & = 0 \end{aligned}$

$\implies \begin{cases} x^2 + x + 1 = 4(x-1) & \implies x^2 - 3x + 5 = 0 & \implies \small \red{\text{No real roots}} \\ x^2 + x + 1 = 9(x-1) & \implies x^2 \blue{ -8}x + 10 = 0 & \implies \small \blue{\text{With real roots}} \end{cases}$

By Vieta's formula sum of real roots is $-(\blue{-8}) = \boxed 8$ .

Squaring and simplifying we get

$x^4-11x^3+39x^2-70x+50=0\implies (x^2-8x+10)(x^2-3x+5)=0$ .

Now in the equation $x^2-8x+10=0$ the discriminant is $8^2-4\times 1\times 10=24>0$ . Hence the roots are real.

In the equation $x^2-3x+5=0$ , the discriminant is $3^2-4\times 1\times 5=-11<0$ . Hence the roots are nonreal.

So, there are only two real roots of the given equation, whose sum is $\boxed 8$ (by Vieta's formula.)