Does |x^2-3x+3|=3 have a solution?

My answer was that there isn't a solution.
Am i right?

The left side can be -3 or 3 since the absolute value of -3 is 3.

Therefore this can be split in to 2 equations:
x^2-3x+3=3
Which simplifies to,
x^2-3x=0

and,

x^2-3x+3=-3
Which simplifies to,
x^2-3x+6=0

Try to solve these and check the discriminant to find the possible solutions.

To determine if the equation |x^2 - 3x + 3| = 3 has a solution, we need to consider two cases: when the expression inside the absolute value is positive and when it is negative.

Case 1: x^2 - 3x + 3 is positive:
To find the solutions, we set x^2 - 3x + 3 equal to 3:
x^2 - 3x + 3 = 3

Simplifying this equation, we have:
x^2 - 3x = 0

Factoring out x, we get:
x(x - 3) = 0

This equation has two solutions: x = 0 and x = 3.

Case 2: x^2 - 3x + 3 is negative:
To find solutions in this case, we set x^2 - 3x + 3 equal to its negative counterpart, -3:
x^2 - 3x + 3 = -3

Simplifying this equation, we have:
x^2 - 3x + 6 = 0

Using the quadratic formula or factoring, we find that there are no real solutions to this equation. Therefore, there are no solutions for this case.

Combining the solutions from both cases, we have x = 0, x = 3 as the solutions for the equation |x^2 - 3x + 3| = 3.

Therefore, your initial answer that there is no solution is incorrect. The correct answer is that the equation has two solutions: x = 0 and x = 3.