急!!!簡單一元二次方程

Questons :
1. 2x^2-9x+9=0
2. 16x^2+8x+1=0
3. x^2+x+4=0
4. x^2-7x+9=0

Answer :
1. 2x^2-9x+9=0
(x-3)(2x-3)=0
x=3 or x=3/2#

2. 16x^2+8x+1=0
(x+1/4)^2=0
x=-1/4 (Repeated)#

3. x^2+x+4=0
By calculate the discriminant, we have,
(1)^2 - 4(1)(4)
=1-16
= -15
Therefore, this quadratic equation has no real root.

4. x^2-7x+9=0
x^2-7x=-9
x^2-7x+(7/2)^2 = -9+(7/2)^2
(x^2-7/2)^2 = 13/4
x-7/2=√(13/4)
x=-7/2±√13 /2
x= (-7±√13)/2 #

2012-09-21 21:41:06 補充：
Sorry, I made mistake. It should be :

3. x^2+x+4=0
By CALCULATING the discriminant, we have,
(1)^2 - 4(1)(4)
=1-16
= -15
Therefore, this quadratic equation has no real root.

1. 2x^2-9x+9=0 (2x+1)(x-9)=0 x=-1/2,9
2.16x^2+8x+1=0 (4x+1)^2 x=-1/4
3. x^2+x+4.......玩呀??????????????????