t^4=6t^2 -t -12 那么 t等于多少?

因为t^4=6t^2 -t -12
所以t^4-6t^2 +t =-12
即t（t^3-6t +1 ）=-12
说明t与（t^3-6t +1 ）积为-12
所以t必为-12的因数
因为-12的因数有1，-1，2，-2，3，-3，4，-4，6，-6，12，-12。
试之，皆不行，即无整数解。
改用图象法，即画出函数
（1）y=t^4
（2）y=6t^2 -t -12 的图象，
交点即是。

该方程没有实根

设f(t) = t^4 - 6t^2 + t + 12 = t^4 - 6t^2 + 9 + t + 3
= [t^2 - 3]^2 + t + 3.
当 t > -3时，f(t) = [t^2 - 3]^2 + t + 3 > 0.
所以，t > -3时，f(t)没有零点。
t f'(t) = 4t^3 - 12t + 1
f''(t) = 12t^2 - 12 = 12(t-1)(t+1)
t 0, f'(t)单调递增。
所以，t 所以，t 所以，t = f(-3) = 36 > 0
所以，t 综合知，f(t）没有零点，也就是说，t^4 = 6t^2 - t - 12没有实根。
没有实数t能使得 t^4 = 6t^2 - t - 12 成立。

牛顿切线法计算方程的近似解吧

和 这个问题的手工计算 计算量可不小呀 我也不想花费那个冤枉时间
下面我使用Matlab帮你计算下吧 很简单的
程序如下：
%by dynamic
%2009.3.5
>> solve('t^4=6*t^2 -t -12')%下面是解析解 够复杂的吧,恩接着后面我将它转化为数值解了 你可以看看 但是没有实根哦
ans =
1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
-1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
-1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
>> vpa(ans,5)
ans =
1.7982-.61035*i
1.7982+.61035*i
-1.7982-.30740*i
-1.7982+.30740*i
>> roots([1 0 -6 1 12])%当然熟悉Matlab的人,可以使用这个命令,求得的结果是一样的
ans =
1.7982 + 0.6103i
1.7982 - 0.6103i
-1.7982 + 0.3073i
-1.7982 - 0.3073i
可以看出该方程没有实根 希望能帮上你