Answer:
2 real solutions
Step-by-step explanation:
[tex]f(x)=(x+3)^2-8\\\\0=(x+3)^2-8\\\\8=(x+3)^2\\\\\pm\sqrt{8}=x+3\\\\x=-3\pm\sqrt{8}[/tex]
Therefore, the function set at [tex]f(x)=0[/tex] has two real solutions (x-intercepts).
You can also check the discriminant of the function by expanding it:
[tex](x+3)^2-8=x^2+6x+9-8=x^2+6x+1\\\\b^2-4ac=6^2-4(1)(1)=36-4=32 > 0[/tex]
Since it's greater than 0, there's two real solutions
