The magnitude of the acceleration and decceleration of an elevator is $4\frac{m}{s^2}$ and its maximum vertical speed is $6\frac{m}{s}$. Find the minimum time (in seconds) such that the elevator goes up and gets to $90\,m$ of height departing from rest and arriving with zero speed.

The alternatives given are:

$\begin{array}{ll}

1.&12.5\,s\\

2.&13.5\,s\\

3.&14.5\,s\\

4.&15.5\,s\\

5.&16.5\,s\\

\end{array}$

What I thought doing here was to use the fact that the combined displacement for going up and deccelerating will add to $90\,m.$

This is summarized as follows:

$y=y_{o}+v_ot+\frac{1}{2}at^2$

The first part reduces to:

$y_{h}=\frac{1}{2}(4)t^2=2t^2$

Then for the second equation is:

$90=y_{h}+6t-\frac{1}{2}4t^2$

But adding the two expressions result into:

$90=6t$

$t=\frac{90}{6}$

Where exactly did I made an error?. Can somebody help me with this?.