How to Re-write completing the square: x2+x+1

Step 1
Remember the formula for the square of a binomial:
${\displaystyle {(a+b)}^2=a^2+2ab+b^2}$
Now, when you see ${\displaystyle x^2+x+1}$, you want to think of ${\displaystyle x^2+x}$ as the first two terms you get in expanding the binomial ${\displaystyle (x+c)2}$ for some c; that is,
${\displaystyle x^2+x+\cdots ={(x+c)}^2}$
Since the middle term should be ${\displaystyle 2cx}$, and you have x, that means that you want ${\displaystyle 2c=1}$, or ${\displaystyle c=\frac{1}{2}}$.
But if you have ${\displaystyle {(x+\frac{1}{2})}^2}$, you get ${\displaystyle x^2+x+\frac{1}{4}}$. Since all you have is ${\displaystyle x^2+x}$, you complete the square by adding the missing ${\displaystyle \frac{1}{4}}$.
Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting ${\displaystyle \frac{1}{4}}$. So:
${\displaystyle x^2+x+2=(x^2+x+\cdots )+1}$
${\displaystyle =(x^2+2\left(\frac{1}{2}\right)x+\cdots )+1}$ figuring out what c is
${\displaystyle =(x^2+2\left(\frac{1}{2}\right)x+{\left(\frac{1}{2}\right)}^2)-\frac{1}{4}+1}$ completing the square
${\displaystyle ={(x+\frac{1}{2})}^2+\frac{3}{4}}$

Step 1
One can rewrite a degree ${\displaystyle n>1}$ polynomial ${\displaystyle f\left(x\right)=x^n+b{x}^{n-1}+\cdots }$ into a form such that its two highest degree terms are ''absorbed'' into a perfect n'th power of a linear polynomial, namely
${\displaystyle f\left(x\right)={(x+\frac{b}{n})}^n-g\left(x\right)}$
where
${\displaystyle g\left(x\right)={(x+\frac{b}{n})}^n-f\left(x\right)}$
has degree ${\displaystyle \le n-2}$
When ${\displaystyle n=2}$ this is called completing the square - esp. when used to solve a quadratic equation. If ${\displaystyle g\left(x\right)=g}$ is constant (as is always true when ${\displaystyle n=2}$) then this yields a closed form for the roots of ${\displaystyle f\left(x\right)}$, namely

$x=\sqrt[n]{g}-\frac{b}{2}$

Step 1 As Arturo points out what you have to observe is the coefficient. $x^2+x+1=x^2+x+\frac{1+3}{4}$ $=x^2+x+\frac{1}{4}+\frac{3}{4}$ $=(x+\frac{1}{2}{)}^2+\frac{3}{4}$ I am sure once you get used to such type of things you shall not have trouble in doing such problems. Solve more problems based on this type. Suppose you have the coefficient of x as a note that $\frac{a^2}{4}$ should be added and subtracted from the constant term. What i mean by this is: Suppose you have something of this type $x^2+ax+b^2$ then you can write this as $(x+\frac{a}{2}{)}^2+b^2-\frac{a^2}{4}$.