习题2.3 (1)x^2+12x+25=0 (2)x^2+4x=10 (3)x^2-6x=11 (4)x^2-2x-4=0 求过程！

解(1):x²+12x+25=0
x²+12x=-25
x²+12x+36=-25+36
(x+6)²=11
x+6=±√11
x+6=√11或x+6=-√11
x1=-6+√11, x2=-6-√11

(2)：x²+4x=10
x²+4x+4=10+4
(x+2)²=14
x+2=±√14
x+2=√14或x+2=-√14
x1=-2+√14, x2=-2-√14

(3)：x²-6x=11
x²-6x+9=11+9
(x-3)²=20
x-3=±√20
x-3=±2√5
x-3=2√5或x-3=-2√5
x1=3+2√5, x2=3-2√5

(4):x²-2x-4=0
x²-2x=4
x²-2x+1=4+1
(x-1)²=5
x-1=±√5
x-1=√5或x-1=-√5
x1=1+√5, x2=1-√5

解：（1）x^2+12x+25=0
x^2+12x+6*6=11
(x+6) ^2=11
x+6= +√11或x+6=-√11
x=+√11-6 或 x=-√11-6

(2)x^2+4x=10
x^2+4x+4=14
(x+2) ^2=14
X=√14-2 或 x=-√14-2
三四同样用配方法去解：

(3)x^2-6x=11
(4)x^2-2x-4=0

x²+12x+25=0
x²+12x=-25
x²+12x+36=-25+36
(x+6)²=11
x+6=±√11
x+6=√11x1=-6+√11, x2=-6-√11