To solve the equations \( 25 - 3x = 100 \) and \( 25 - 3x^2 = 100 \), let's tackle them one at a time.

### Solving the first equation: \( 25 - 3x = 100 \)

1. Subtract 25 from both sides:
\[
-3x = 100 - 25
\]
\[
-3x = 75
\]

2. Divide both sides by -3:
\[
x = \frac{75}{-3} = -25
\]

So, the solution to the first equation is:
\[
x = -25
\]

### Solving the second equation: \( 25 - 3x^2 = 100 \)

1. Subtract 25 from both sides:
\[
-3x^2 = 100 - 25
\]
\[
-3x^2 = 75
\]

2. Divide both sides by -3:
\[
x^2 = \frac{75}{-3} = -25
\]

At this point, we see that \( x^2 = -25 \) does not have real solutions, because the square of a real number cannot be negative. Thus, we can conclude that:

- The equations yield \( x = -25 \) for the first equation.
- The second equation has no real solutions.

If you're looking for complex solutions for the second equation, we can find them:

3. Taking the square root of both sides:
\[
x = \pm \sqrt{-25} = \pm 5i
\]

Thus, the solutions for the second equation are:
\[
x = 5i \quad \text{and} \quad x = -5i
\]

To summarize:
- First equation: \( x = -25 \)
- Second equation: \( x = 5i \) and \( x = -5i \) (complex solutions).