Answer:
[tex]F(x)=2\left(x+\dfrac{13}{4}\right)^2-\dfrac{9}{8}.[/tex]
Step-by-step explanation:
Consider the function [tex]F(x)=2x^2+13x+20.[/tex]
Rewrite it as follows:
[tex]F(x)=(2x^2+13x)+20=2\left(x^2+\dfrac{13}{2}x\right)+20=2\left(x^2+\dfrac{13}{2}x+\left(\dfrac{13}{4}\right)^2-\left(\dfrac{13}{4}\right)^2\right)+20=2\left(x^2+\dfrac{13}{2}x+\dfrac{169}{16}\right)-2\cdot \dfrac{169}{16}+20.[/tex]
Since
[tex]x^2+\dfrac{13}{2}x+\dfrac{169}{16}=\left(x+\dfrac{13}{4}\right)^2,[/tex]
you get
[tex]F(x)=2\left(x+\dfrac{13}{4}\right)^2+20-\dfrac{169}{8}=2\left(x+\dfrac{13}{4}\right)^2-\dfrac{9}{8}.[/tex]
