## anonymous 5 years ago need to find 2 solutions to this equation

1. anonymous

$(3x+8)^{2}-4(3x+8)-77=0$

2. anonymous

Let k = (3x+8) So the above equation becomes $k^2 - 4k - 77 = 0$ Solve the quadratic for two solutions of k, then replace k in those two equations with 3x+8 and solve for your two x solutions.

3. anonymous

ok thanks, I will try that out. Not confident I'll get it right so you can correct me if it's wrong :)

4. anonymous

i'm not sure if i'm doing this step correct. I got -76 for one..?

5. anonymous

$k = \frac{4 \pm \sqrt{16 + 4*77}}{2}$ $k = 2 \pm \sqrt{324}$ $k = 2 \pm 18$ $k = -16 \text{ or } k = 20$ Therefore $3x+8 = -16$ or $3x+8 = 20$

6. anonymous

So working with the first equation: $3x + 8 = -16$ $3x = -24$ $x = -8$

7. anonymous

Working the second one: $3x+8 = 20$ $3x = 12$ $x = 4$

8. anonymous

So our two solutions are $$x \in \{-8,4\}$$