I don’t know what you’re asking because you haven’t actually asked a question.
So I just assume you might want to know what values x can take so that the square root can be evaluated.
67+4x-2x²=67-2(x²-2x)=67-2(x²-2x+1)+2=
69-2(x²-2x+1)=69-2(x-1)².
When this expression is greater than or equal to zero, the square root can be evaluated.
So 69-2(x-1)²≥0, 69≥2(x-1)². So (x-1)²≤69/2.
Since (x-1)²=(1-x)², so that x-1≤√(69/2) or 1-x≤√(69/2), meaning x≤1+√(69/2) or x≥1-√(69/2) or 1-√(69/2)≤x≤1+√(69/2) or about -4.87367≤x≤6.87367 or [-4.87367,6.87367] is the approximate interval for which y exists.
If you were to graph the equation, this would show a curve lying above the x-axis between the calculated interval.