I have this basic problem:
In a farm, $X$ animals are added to the farm. These animals gain weight according to the equation: $500 - 2X$ gr. Which interval of animals can the farm take, if the total weight gain is greater than $30,600$ Kg?
Original (Spanish - Español):
En un criadero de cuyes se integran $x$ cuyes, si se tiene presente que los cuyes ganan peso en promedio de $(500 - 2x)$ gramos. ¿Qué intervalo de cuyes puede aceptar esta granja si la ganancia total de peso de los cuyes es mayor a $30 600$ Kg?
Step 1:
Total animals: $x$.
Weigthgain = $(500 - 2x)$ gr.
TotalWeightgain > $30 600(1000)$ converting kg to gr.
Step 2:
$$x(500 - 2x) > 30600000$$
Step 3:
$$0 > 2x^2 - 500x + 30 600 000$$
Step 4:
$$ x = \frac{-(-500) +- \sqrt[2]{(-500)^2 - 4(2)(30 600 000)}}{2(2)}$$
But as you can see, the sqrt is negative. So It does not exist. What should be the next step?
Assuming that you've transcribed and translated the question correctly, then your math is correct, and indeed no solution exists. So the next step would be to state that it is impossible for the farm to satisfy these conditions (there is no interval, or an empty interval, of solutions).
However, I suspect that there's an error in the units; likely they were intended to be the same (not grams in one place, kg in another). If we graph $y = x(500 - 2x)$, then we see that the maximum point occurs at (125, 31,250). While that's below 30,600,000 by some 3 orders of magnitude, it is slightly above 30,600, and so you will have a nonempty solution interval if the requirements were in matching units. So that's likely what was intended.
(Another slight correction to what you wrote is that $x$ isn't the total animals on the farm, it's just the added animals. I'm assuming that only these added animals gain weight, not the ones originally present, which are an unknown quantity.)