Question

2x^2-9x+9=0 complete square

Answer 1

this can be factored

Answer 2

The factor 2 makes it a little harder:
\[2x^2-9x+9=2(x^2-\frac{ 9 }{ 2 } x)+9=2((x-\frac{ 9 }{ 4 })^2-\frac{ 81 }{ 16 })+9\]Now work out the outer brackets:\[2(x-\frac{ 9 }{ 2 })^2-\frac{ 81 }{ 8 }+\frac{ 72 }{ 8 }=2(x-\frac{ 9 }{ 2 })^2-\frac{ 9 }{ 8 }=0\]Because of the "=0" part, you probably have to use this form to solve for x. Let me know if you cannot do that!