[tex]2x^2+6x+11=2(x^2+3x)+11=2(x^2+2x\cdot1.5)+11\\\\=2(x^2+2x\cdot1.5+1.5^2-1.5^2)+11=2[(x+1.5)^2-2.25]+11\\\\=2(x+1.5)^2-4.5+11=2(x+1.5)^2+6.5[/tex]
Other method:
[tex]f(x)=ax^2+bx+c\to f(x)=a(x-h)^2+k\\\\h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}\\\\f(x)=2x^2+6x+11\\\\a=2;\ b=6;\ c=11\\\\h=\dfrac{-6}{2\cdot2}=-\dfrac{6}{4}=-\dfrac{3}{2}=-1.5\\\\k=f(1.5)=2(-1.5)^2+6\cdot(-1.5)+11=2\cdot2.25-9+11=6.5\\\\2x^2+6x+11=2(x+1.5)^2+6.5[/tex]
