Answer:
[tex](x-\frac{1}{2})^2=\frac{81}{4}[/tex]
Step-by-step explanation:
Hi there!
[tex]x^2-x-20=0[/tex]
This equation is written in the form [tex]ax^2+bx+c=0[/tex]. First, use partial factoring:
[tex](x^2-x)-20=0[/tex]
For x^2-x, the b value is -1 in [tex]ax^2+bx+c[/tex]. To complete the square, take the square of half of 1 and add it in the parentheses as the c value:
[tex](x^2-x+\frac{1}{2}^2 )-20=0[/tex]
However, when adding values to one side of the equation, we must to the same to the other side:
[tex](x^2-x+\frac{1}{2}^2 )-20=\frac{1}{2}^2[/tex]
Complete the square:
[tex](x-\frac{1}{2})^2-20=\frac{1}{4}[/tex]
Move 20 to the other side
[tex](x-\frac{1}{4})^2=\frac{1}{4}+20\\(x-\frac{1}{2})^2=\frac{81}{4}[/tex]
I hope this helps!
