Since the first term is the square of 1/2x and the second term is twice the product of 1/2x and 1, we want to complete the following square:
[tex]\left(\dfrac{1}{2}x+1\right)^2=\dfrac{1}{4}x^2+x+1[/tex]
To do so, we add 3/4 to both sides:
[tex]\dfrac{1}{4}x^2+x+\dfrac{1}{4}=0 \iff \dfrac{1}{4}x^2+x+1=\dfrac{3}{4}[/tex]
So, we can now write the equation as
[tex]\left(\dfrac{1}{2}x+1\right)^2=\dfrac{3}{4}[/tex]
And then continue as usual:
[tex]\dfrac{1}{2}x+1\right=\pm\dfrac{\sqrt{3}}{2}[/tex]
[tex]\dfrac{1}{2}x=\pm\dfrac{\sqrt{3}}{2}-1[/tex]
[tex]x=\pm\sqrt{3}-2[/tex]
