Question 1193299

{{{w(2w+1)=15}}}.....first multiply 

{{{w*2w+w*1=15}}}

{{{2w^2+w=15}}}

{{{2w^2+w-15=0}}}........factor, write {{{w}}} as {{{6w-5w}}}

{{{2w^2+6w-5w-15=0}}}....group

{{{(2w^2+6w)-(5w+15)=0}}}

{{{2w(w+3)-5(w+3)=0}}}

{{{(w + 3)(2w - 5) = 0}}}

solutions:

if {{{(w + 3) = 0}}} =>{{{w=-3}}}

if {{{(2w - 5) = 0}}} =>{{{w=5/2}}}