Answer:
2 (real) solutions.
Explanation:
A quadratic always has two solutions, whether they are real or complex.
Sometimes the solution is complex, involving complex numbers (2 complex), sometimes they are real and distinct (2 real), and sometimes they are real and coincident (still two real, but they are the same).
In the case of
x^2+3x = 3, or
x² + 3x -3 = 0
we apply the quadratic formula to get
x = (-3 +/- sqrt(3^2+4(1)(3))/2
to give the two solutions
{(sqrt(21)-3)/2, -(sqrt(21)+3)/2,}
both of which are real.