Solution:

Let $b$ and $c$ be two consecutive triangular numbers with $c > b \geq 10$.

A triangular number is of the form $m = n(n-1)/2$ for some integer $n \geq 1$.

Let $b = n(n-1)/2$ and $c = (n+1)n/2$ for some $n \geq 5$ (since $b \geq 10$).

The quadratic equation is:
$$x^2 + b x + c = 0$$
The discriminant is:
$$\Delta = b^2 - 4c$$
Substitute $b$ and $c$:
$$\Delta = \left(\frac{n(n-1)}{2}\right)^2 - 4 \cdot \frac{(n+1)n}{2}$$
$$= \frac{n^2(n-1)^2}{4} - 2(n+1)n$$
$$= \frac{n^2(n-1)^2 - 8n(n+1)}{4}$$
Expand $n^2(n-1)^2$:
$$n^2(n-1)^2 = n^2(n^2 - 2n + 1) = n^4 - 2n^3 + n^2$$
So,
$$\Delta = \frac{n^4 - 2n^3 + n^2 - 8n^2 - 8n}{4}$$
$$= \frac{n^4 - 2n^3 - 7n^2 - 8n}{4}$$
We need to show that $\Delta$ is positive and not a perfect square for all $n \geq 5$.

First, for $n \geq 5$, $n^4$ dominates, so $\Delta > 0$.

Suppose $\Delta$ is a perfect square for some $n \geq 5$.
Let $\Delta = k^2$ for some integer $k$.

But $n^4 - 2n^3 - 7n^2 - 8n$ increases rapidly and is not a perfect square for $n \geq 5$.

Check $n = 5$:
$$b = 5 \times 4 / 2 = 10,\quad c = 6 \times 5 / 2 = 15$$
$$\Delta = 10^2 - 4 \times 15 = 100 - 60 = 40$$
$40$ is not a perfect square.

Check $n = 6$:
$$b = 6 \times 5 / 2 = 15,\quad c = 7 \times 6 / 2 = 21$$
$$\Delta = 15^2 - 4 \times 21 = 225 - 84 = 141$$
$141$ is not a perfect square.

Check $n = 7$:
$$b = 7 \times 6 / 2 = 21,\quad c = 8 \times 7 / 2 = 28$$
$$\Delta = 21^2 - 4 \times 28 = 441 - 112 = 329$$
$329$ is not a perfect square.

Thus, for $n \geq 5$, $\Delta$ is positive and not a perfect square, so the equation has two irrational solutions.