Help, i got my answer to y = 3 but it says the soloution also could be -1/4 how?

I don't know if this explains why but... it's because the answer has separate solutions, the two solutions can be written as \({(8y-11)^2}=169 \\ \downarrow Square-root \space both \space of \\ the \space equation \\ 8y-11=±13 \\ \downarrow Separate \space the \space solutions \\ 8y-11=13 \\ or \\ 8y-11=-13 \)

and this is why there are two answers.

\((8y-11)^2 = 169\\ 64y^2 - 176y + 121 = 169\\ 64y^2 - 176y -48 = 0\\ 16y^2 - 44y - 12 = 0\\ 4y^2 - 11y - 3 = 0\\ (y-3)(4y+1) = 0\\ y = 3 \text{ or }y = -\dfrac{1}{4}\)

I think what you have done is just taking square roots for both sides, but this method cannot always give all the answers to a polynomial equation.(including quadratics)

Or you could simply say

 

8y - 11 = 13   and   8y - 11 = -13   are the two possibilities

 

From one you get y = (13 + 11)/8 → 3

 

From the other you get y = (-13 + 11)/8 → -1/4

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There is nothing wrong with taking the square root, it is afterall where, after completion of the square, the formula for the solution of a quadratic comes from.

Taking the square root, 8y - 11 = 13 from which y = 3, or, 8y - 11 = -13 from which y = -1/4.