You Can't Solve This

Duh.....I should have seen right off that 3 was a "root"......over-reliance on technology is sometimes crippling......LOL!!!!!

Well...maybe we can.....

y=(-6/y)²-1         simpify

y = 36/y^2 - 1    multiply through by y^2

y^3 = 36 - y^2

And using the onsite solver, we have one real solution and two non-real solutions...

$${{\mathtt{y}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{36}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{2}}\\ {\mathtt{y}} = {{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{2}}\\ {\mathtt{y}} = {\mathtt{3}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{2.828\: \!427\: \!124\: \!746\: \!190\: \!1}}{i}\\ {\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.828\: \!427\: \!124\: \!746\: \!190\: \!1}}{i}\\ {\mathtt{y}} = {\mathtt{3}}\\ \end{array} \right\}$$

y=(-6/y)²-1

$$\\y=\left(\frac{-6}{y}\right)^2-1\\\\ y=\frac{36}{y^2}-1\\\\ y^3=36-y^2\qquad y\ne 0\\\\ y^3+y^2-36=0\\\\$$

now I am going to use reaminder theorum to look for a factor

If y=3 then  27+9-36=0  great (y-3) is a factor and 3 is a root

$$\\(y^3+y^2-36)\div (y-3)=y^2+4y+12 \qquad $I used polynomial division to get this$\\\\ so\\ y^3+y^2-36=(y-3)(y^2+4y+12)\\\\\\ $Now find roots of $y^2+4y+12\\\\ \triangle = 16-48<0\qquad $therefore there are no real roots to this$\\\\ so\\ $The only real solution is $y=3$$

check

http://www.wolframalpha.com/input/?i=y^3%2By^2-36%3D0

Yes that is true :))