Algebra

First, let's move all the terms to one side and combine all like terms. 

We have the inequality

\(4t^2-48+23t\le \:0\)

Now, let's complete the square for t. We get

\(4\left(t+\frac{23}{8}\right)^2-\frac{1297}{16}\le \:0\)

This format allows us to isolate t in a much simpler manner. Now, let's start the process. We get

\(4\left(t+\frac{23}{8}\right)^2\le \:\frac{1297}{16}\)

\(\left(t+\frac{23}{8}\right)^2\le \:\frac{1297}{64}\)

Now, square rooting both sides, it gives us two parameters to focus on for t. We get

\(-\sqrt{\frac{1297}{64}}\le \:t+\frac{23}{8}\le \:\sqrt{\frac{1297}{64}}\\ \frac{-\sqrt{1297}-23}{8}\le \:t\le \:\frac{\sqrt{1297}-23}{8}\)

Now, the final step is to write this in interval notation. We finally get that t is \([\frac{-\sqrt{1297}-23}{8},\frac{\sqrt{1297}-23}{8}]\)

Thanks! :)