## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$$t=15$$
We solve the equation using the rules of exponents and radicals. Note, if we wanted to check our solution, we could plug the value we get back into the original equation to see if it yields a true statement. Solving the equation, we find: $$\left(7+\sqrt{2t-5}\right)^2=\left(3\sqrt{t+1}\right)^2\\ 2t+14\sqrt{2t-5}+44=9t+9\\ \left(14\sqrt{2t-5}\right)^2=\left(7t-35\right)^2\\ 392t-980=49t^2-490t+1225\\ t=15$$ Note, $t=3$ is not a solution.