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2015-06-04T00:15:58+08:00

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Yes,it's a quadratic ecuation because it has the general form of a second grade ecuation which is:

ax^2+bx+c

We have the unknown therm raised at the second power x^2 = ax^2   => a=1
So,we would solve the quadratic ecuation:

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}


We do the sign table
      x       |-                   \frac{5}{2}}                   +   
x²-5x+10|+∞      ↓          \frac{15}{4}        ↑        +∞

S=(-\infty; \frac{5}{2}] \ \cup \ [\frac{5}{2};\infty)
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