(2t+3)平方 · (2t+3)=6t+9 求t解 急用

(2t+3)² * (2t+3) = 6t+9
(2t+3)² * (2t+3) = 3(2t+3)
(2t+3)² * (2t+3) - 3(2t+3) = 0
(2t+3) [(2t+3)² - 3] = 0
(2t+3)(2t+3+√3)(2t+3-√3) = 0
t1=-(3+√3)/2，t2=-3/2，t3=-(3-√3)/2

(2t+3)^2×(2t+3)=3(2t+3)
(2t+3)^2=3
2t+3=±根号3
t=(3±根号3)/2

(2t+3)平方x(2t+3)=3(2t+3)
(2t+3)平方=3
2t+3=正负根号3
t=2分之正负根号3-3
当2t-3=0时 方程也成立 即 t=2分之3

(2t+3)^3=3*(2t+3)
所以t= -3/2或者t=(正负根号3-3)/2