Solve `x^2-3x+4=0` by completing the square:

`x^2-3x+4=0` subtract 4 from both sides

`x^2-3x=-4` Now add `(-3/2)^2` to both sides

`x^2-3x+9/4=-4+9/4`

The left hand side is now a perfect square trianomial; `a^2-2ab+b^2=(a-b)^2` . Here a=x and `b=3/2` so

`(x-3/2)^2=-7/4` Take the square root of both sides:

`x-3/2=+-sqrt(-7/4)`

`x=3/2+-sqrt(7)/2 i` where `i=sqrt(-1)`

...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Solve `x^2-3x+4=0` by completing the square:

`x^2-3x+4=0` subtract 4 from both sides

`x^2-3x=-4` Now add `(-3/2)^2` to both sides

`x^2-3x+9/4=-4+9/4`

The left hand side is now a perfect square trianomial; `a^2-2ab+b^2=(a-b)^2` . Here a=x and `b=3/2` so

`(x-3/2)^2=-7/4` Take the square root of both sides:

`x-3/2=+-sqrt(-7/4)`

`x=3/2+-sqrt(7)/2 i` where `i=sqrt(-1)`

-----------------------------------------------------------

If you meant to write in vertex form we follow a similar procedure:

`y=x^2-3x+4` Add and subtract `9/4`

`y=x^2-3x+9/4-9/4+4`

`y=(x-3/2)^2+7/4` which is in vertex form.

***************************************************

Why `(-3/2)^2` ? We want to add a number k such that `x^2-3x+k` is a perfect square trianomial. Then we need `2(1)(sqrt(k))=-3==>sqrt(k)=-3/2==>k=(-3/2)^2`