Question 1186014
{{{w}}}, the width
{{{2w+5}}}, the length
{{{w(2w+5)=52}}}, the area


Apply quadratic equations?

More than one way to do.

Try maybe  ((((2w)(2w+5)=2*52}}}, or {{{(2w)(2w+5)=104}}}; and MAYBE you might find two factors differing by five, whose product is 104.


Or try for factoring or general solution:
{{{2w^2+5w-52=0}}}

{{{w=(-5+sqrt(5^2-4*2*(-52)))/(2*2)}}}

{{{w=(-5+sqrt(25+416))/(2*2)}}}

{{{w=(-5+sqrt(441))/4}}}

{{{w=(-5+21)/4}}}

{{{w=16/4}}}

{{{highlight(w=4)}}}