Question 866123
{{{w}}}= width of the rectangle (in whatever units of length they use)
{{{2w}}}= twice the width
{{{2w-5}}}= length of the rectangle (it is 5 less than twice the width)
The area of the rectangle is length times width, so it is
{{{(2w-5)w=63}}}
That is the equation to solve, but we re-arrange it.
{{{(2w-5)w=63}}}
{{{2w^2-5w=63}}}
{{{2w^2-5w-63=0}}}
Now it looks like a standard {{{ax^2+bx+c=0}}} quadratic equation and we can solve it using the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ,
except that instead of {{{x}}} we have {{{w}}} , and we have
{{{a=2}}} , {{{b=-5}}} , and {{{c=-63}}} , so
{{{w = (-(-5) +- sqrt((-5)^2-4*2*(-63)))/(2*2)=(5 +- sqrt(25+504))/4=(5 +- sqrt(529))/4=(5 +- 23)/4}}}
The two solutions for {{{2w^2-5w-63=0}}} are
{{{w=(5 + 23)/4=28/4=highlight(7)}}} and
(5 - 23)/4=-18/4=-9/2}}} ,
but since the width of the rectangle must be a positive number,
we must discard {{{w=-9/2}}} , and the solution is
{{{highlight(w=7)}}} , along with
{{{length=2w-5=2*7-5=14-5=highlight(9)}}}
 
The same {{{2w^2-5w-63=0}}} quadratic equation can be solved by other means.
Ypu could "complete the square" (not so easy),
or you could solve it by factoring {{{2w^2-5w-63}}} (if you are good at factoring).
Factoring:
{{{2w^2-5w-63=(2w^2-14w)+(9w-63)=2w(w-7)+9(w-7)=(2w+9)(w-7)}}}
so {{{2w^2-5w-63=0}}} can be written as {{{(2w+9)(w-7)=0}}} ,
which is true if {{{system(2w+9=0,"or",w-7=0)}}} ,
and those two options lead us to the solutions
{{{w=-9/2, "or", w=7)}}} found above.
 
Completing the square:
{{{2w^2-5w-63=0}}}<-->{{{2w^2-5w=63}}}
dividing both sides of the equal sign by 2,
{{{w^2-(5/2)w=63/2}}}
{{{w^2-(5/2)w+(5/4)^2=63/2+(5/4)^2}}}
{{{(w-5/4)^2=63/2+25/16}}}
{{{(w-5/4)^2=504/16+25/16}}}
{{{(w-5/4)^2=529/16}}}
So {{{system(w-5/4=sqrt(529/16)=sqrt(529)/sqrt(16)=23/4,"or",w-5/4=-sqrt(529/16)=-sqrt(529)/sqrt(16)=-23/4)}}}
which means that
{{{system(w=23/4+5/4=28/4=7,"or",w=-23/4+5/4=-18/4=-9/2)}}}