Question 936200

{{{5y^2+5y-25y-25=0}}}...as you can see, all terms are divisible by {{{5}}}; so, we will simplify it first

{{{5y^2/5+5y/5-25y/5-25/5=0}}}

{{{y^2+y-5y-5=0}}}

{{{y^2-4y-5=0}}}...factor completely; first write {{{-4y}}} as {{{y-5y}}}

{{{y^2+y-5y-5=0}}}...group

{{{(y^2+y)-(5y+5)=0}}}

{{{y(y+1)-5(y+1)=0}}}

{{{(y-5)(y+1)=0}}}

so, solutions are:

if {{{(y-5)=0}}} then {{{highlight(y=5)}}}

if {{{(y+1)=0}}} then {{{highlight(y=-1)}}}