Question 172893
The same way you would solve any quadratic.


First put it in standard form:  {{{y^2-y-30=0}}}


Fortunately, this one factors because {{{-6*5=30}}} and {{{-6+5=-1}}}, so


{{{(y-6)(y+5)=0}}}


{{{y-6=0}}} => {{{y=6}}}

or

{{{y+5=0}}} => {{{y=-5}}}