A blue ball is thrown upward with an initial speed of 21.2 m/s, from a height of 0.9 meters above the ground. 2.6 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10.8 m/s from a height of 25.4 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

1. How long after the blue ball is thrown are the two balls in the air at the same height?
In this problem, what variable is equal to each other, because I have tried many ways and it kept telling me it's wrong.

You're clearly on the right track, knowing that two variables MUST be equal to each other to find the answer.

In both cases you can compute the ball's position with the equation

x = x_0 + v_0 \cdot t + \frac{1}{2}a \cdot t^2

I suspect you're on the right track, setting the two equations for x equal to each other (so that you can solve for t).

x_{blue}=x_{red}

However, there's one part of this that's a little tricky: The red ball begins it's flight 2.6 seconds after the blue ball. Since both equations have to be referenced to the same time, t, you need to replace all of the t's in the equation for the red ball with (t - 2.6). Then when you set the positions of the two balls to be equal to each other, you should be able to solve for t, then plug back into either of the equations to solve for the height, x.