need to find 2 solutions to this equation

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need to find 2 solutions to this equation

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[(3x+8)^{2}-4(3x+8)-77=0\]
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.
ok thanks, I will try that out. Not confident I'll get it right so you can correct me if it's wrong :)

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i'm not sure if i'm doing this step correct. I got -76 for one..?
\[k = \frac{4 \pm \sqrt{16 + 4*77}}{2}\] \[k = 2 \pm \sqrt{324}\] \[k = 2 \pm 18\] \[k = -16 \text{ or } k = 20\] Therefore \[3x+8 = -16\] or \[3x+8 = 20\]
So working with the first equation: \[3x + 8 = -16\] \[3x = -24\] \[x = -8\]
Working the second one: \[3x+8 = 20\] \[3x = 12\] \[x = 4\]
So our two solutions are \(x \in \{-8,4\}\)

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