Yes,it's a quadratic ecuation because it has the general form of a second grade ecuation which is:
[tex]ax^2+bx+c[/tex]
We have the unknown therm raised at the second power x^2 = ax^2 => a=1
So,we would solve the quadratic ecuation:
[tex]x^2-5x+10=0 \\\\ a=1 \\ b=-5 \\ c=10 \\\\ \Delta= b^2-4ac= (-5)^2-4*1*10= 25-40\to \boxed{-15} \ \textless \ 0 \\ \Delta \ \textless \ 0 \\ a\ \textgreater \ 0 \\\\ -\frac{b}{2a}\to \frac{5}{2} \\\\ -\frac{\Delta}{4a}\to\frac{15}{4}[/tex]
We do the sign table
x |-∞ [tex]\frac{5}{2}}[/tex] +∞
x²-5x+10|+∞ ↓ [tex]\frac{15}{4}[/tex] ↑ +∞
[tex]S=(-\infty; \frac{5}{2}] \ \cup \ [\frac{5}{2};\infty)[/tex]
