(x+3/t = t^2
t^3 = t+3
t^3 - t - 3 = 0
This is not easily done with "pencil and paper"
You will need a program that solves cubics on your calculator or computer.
You could use a method called Newton's Method,
but this webpage will solve a cubic for you
http://www.1728.com/cubic.htm
enter 1,0,-1,-3 for A,B,C, and D
X+3 over t =t squared
2 answers
I assume that the original equation should read (t+3)/t = t², not (x+3)/t = t², yes?
You can actually do it quite quickly by iteration if you've got access to a calculator with a cube-root function (or logs and antilogs). Rearrange it so that it reads:
t = (t+3)^(1/3)
Put t=1 and evaluate the right-hand side. You'll get the cube root of 4, i.e. 1.5874. Now feed 1.5874 into the equation, and you'll get 1.6616. Feed that in and you'll get 1.6705. Repeat the operation a few more times and the answer will converge to the answer you're looking for.
You can actually do it quite quickly by iteration if you've got access to a calculator with a cube-root function (or logs and antilogs). Rearrange it so that it reads:
t = (t+3)^(1/3)
Put t=1 and evaluate the right-hand side. You'll get the cube root of 4, i.e. 1.5874. Now feed 1.5874 into the equation, and you'll get 1.6616. Feed that in and you'll get 1.6705. Repeat the operation a few more times and the answer will converge to the answer you're looking for.