1、x²-10x+21=0
(x-3)(x-7)=0
x1=3,x2=7
2、原方程等于：
X^2 - 2*(1/2)*X + (1/2)^2 - (1/2)^2 - 1 = 0
=> (X - 1/2)^2 - 5/4 = 0
=> (X - 1/2)^2 = 5/4
=> X - 1/2 = ±√(5/4)
=> X1 = √(5/4) + 1/2
X2 = -√(5/4) + 1/2
=> X1 = (1+√5)/2
=> X2 = (1-√5)/2
3、一元二次方程式ax^2+b+c=0
檢驗(yàn)有沒有解:
b^2-4ac>0本題中有兩個(gè)不相等的實(shí)數(shù)解
3x^2+6x-4=0
x^2+2x-4/3=0
x^2+2x+1-1-4/3=0
(x+1)^2-7/3=0
(x+1)^2=7/3
(x+1)=±√21/3
x=±√21/3-1
4、3x（x+1）=3x+3
3x^2+3x=3x+3
3x^2+3x-3x-3=0
3(x^2-1)=0
x^2-1=0
x^2=1
x=±1
5、 4x^2-4x+1=x^2+6x+9
3x^2-10x-8=0
x^2-10x/3-8/3=0
(x-10/6)^2-25/9-8/3=0(x-10/6)^2=49/9
x-5/3=±7/3
x=±7/3+5/3
x=4或 -2/3
6、7x²-√6x-5=0
x=[-b±√（b²-4ac）]/2a
x=(√6±√146)/14