How many solutions with given condition?

$a+b+5c=0 \Leftrightarrow b= -(a+5c)$

Examine $\Delta = b^2-4ac$ :

$\Delta = [-(a+5c)]^2 -4ac = (a^2+10ac+25c^2)-4ac = (a^2+6ac + 9c^2) + 16c^2 = (a+3c)^2+16c^2 \geq 0$

Therefore, the quadratic equation has at least 1 distinct real solution.

Examine the scenario where the quadratic equation has only 1 distinct real solution: (or $\Delta = 0$ )

$\Rightarrow \left\{\begin{matrix} a+3c=0\\ c=0 \end{matrix}\right.$

$\Leftrightarrow a=c=0$

However, since $ax^2+bx+c=0$ is a quadratic equation, $a \neq 0$ .

Hence, the quadratic equation has $\boxed{2}$ distinct real solutions.

This is equivalent of proving that the discriminant (in $x$ ) is strictly positive. That is, the number of distinct real solutions to the quadratic equation in question is $\boxed2.$

Claim: The discriminant is strictly positive.

Proof: Suppose not. That is, suppose that it is non-negative, then it satisfies the inequality: $b^2-4ac\leqslant 0$ . With $a + b+ 5c=0,$ we get $(a+5c)^2 - 4ac \leqslant 0 \quad\implies \quad (a+3c)^2 + (4c)^2 \leqslant 0 \quad \implies \quad a+3c = 4c = 0 \implies a=c=b=0$ But this means we don't have a quadratic equation to begin with. A contradiction!