(y-2x)^2 + (y+x^2)^2 = 0

To solve the equation \((y-2x)^2 + (y+x^2)^2 = 0\), we need to analyze what this means for the two squared terms.

Since both terms in the sum are squares, each term must be equal to zero for the sum to be zero:

1. \( (y - 2x)^2 = 0 \)
2. \( (y + x^2)^2 = 0 \)

From the first equation:

\[ y - 2x = 0 \implies y = 2x \]

From the second equation:

\[ y + x^2 = 0 \implies y = -x^2 \]

Now, we have two expressions for \(y\):

1. \(y = 2x\)
2. \(y = -x^2\)

Next, we can set these two expressions equal to each other to find \(x\):

\[ 2x = -x^2 \]

Rearranging gives:

\[ x^2 + 2x = 0 \]

Factoring this equation:

\[ x(x + 2) = 0 \]

This gives us the solutions:

1. \(x = 0\)
2. \(x = -2\)

Now, we can substitute these \(x\) values back into either equation to find the corresponding \(y\) values.

1. For \(x = 0\): \[ y = 2(0) = 0 \] Thus, one solution is \((0, 0)\).

2. For \(x = -2\): \[ y = 2(-2) = -4 \] Thus, another solution is \((-2, -4)\).

In conclusion, the solutions to the equation \((y-2x)^2 + (y+x^2)^2 = 0\) are:

\[ (0, 0) \quad \text{and} \quad (-2, -4) \]