Answer:
[tex]x_1=2; x_2=-1[/tex]
Step-by-step explanation:
The square term is [tex]x^2[/tex] so we expect the complete square to be of the form [tex](x-a) ^2=x^2+2ax+a^2[/tex].
Let's compare the first degree terms now: we have [tex]2ax[/tex] on one side and [tex]-1x[/tex] on the other. That would make [tex]a=- \frac12[/tex]. At this point we have still the second squared term, so we add and subtract [tex]\frac14 = (\frac12)^2[/tex]
Our equation becomes:
[tex]x^2-x+\frac14 -\frac14-2 =0\\(x-\frac12)^2-\frac94=0[/tex]
(the [tex]\frac94[/tex] term comes from rewriting [tex]2= \frac84[/tex] and adding the terms out of the bracket).
At this point it's just solving the equation:
[tex](x-\frac12)^2 = \frac94 \rightarrow x-\frac12=\pm\frac32\\x=\frac12\pm\frac32\\x_1=\frac42 = 2; x_2=\fra-\frac22=-1[/tex]
