用配方法解方程-x²+7x-3=0 ,（2y+1）²+15=8（2y+1）.
问题描述：
用配方法解方程-x²+7x-3=0 ,（2y+1）²+15=8（2y+1）.
答
（1）-x^2+7x-3=0 ,
x^2-7x+3=0 ,
x^2-7x = -3 ,
x^2-7x+(7/2)^2 = -3+(7/2)^2 ,
(x-7/2)^2 = 37/4 ,
x - 7/2 = ±√37/2 ,
所以 x1 = (7-√37)/2 ,x2 = (7+√37)/2 .
（2）(2y+1)^2 +15 = 8(2y+1) ,
(2y+1)^2-8(2y+1)+15 = 0 ,
(2y+1)^2-8(2y+1) = -15 ,
(2y+1)^2-8(2y+1)+4^2 = -15+4^2 ,
[(2y+1)-4]^2 = 1 ,
(2y-3)^2 = 1 ,
2y-3 = ± 1 ,
因此,由 2y-3 = - 1 得 y = 1 ；由 2y-3 = 1 得 y = 2 .