Question 730838
You are on the right path and just need to solve  {{{2x^2 + x -55 = 0 }}}
There are easier quadratic equations

 
I know of 3 ways to solve quadratic equations like {{{2x^2 + x -55 = 0 }}}
Factoring will work if the answers are rational numbers.
Completing the square will always work.
Using the quadratic formula will always work too.
 
FACTORING:
When the leading coefficient is not 1 or -1, factoring is a little harder.
In this case, you must look for pairs of factors of {{{2*55=110}}} .
Giving a negative sign to one factor and a positive sign to the other, they must add up to the coefficient of the term in x, 1.
The number 110 can be written as 4 different products:
{{{110=1*110}}}
{{{110=2*55}}}
{{{110=5*22}}}
{{{110=10*11}}}
The last one is the one that works because {{{11-10=1}}}
So we use 11 and -10 as coefficients of x and write {{{2x^2 + x -55}}} as
{{{2x^2 + 11x -10x -55}}}
Then we factor by grouping, like this:
{{{2x^2 + 11x -10x -55=(2x^2 -10x) + (11x -55)=2x(x-5)+11(x-5)=(2x+11)(x-5)}}}
So, since {{{2x^2 + x -55=(2x+11)(x-5)}}} we re-write the equation as
{{{(2x+11)(x-5)=0}}} and find the solutions that make
{{{2x+11=0}}} --> {{{x=-11/2}}} and
{{{x-5=q}}} --> {{{x=5}}}
Since x must be positive to be the width of a rectangle, the only solution is {{{x=5)}}}.
 
COMPLETING THE SQUARE:
{{{2x^2 + x = 55 }}} --> {{{x^2+x/2=55/2}}} dividing both sides by 2
{{{x^2+x/2}}} is part of {{{x^2+x/2+1/16=(x+1/4)^2}}} so if we add {{{1/16}}} to both sides we "complete the square:
{{{x^2+x/2=55/2}}} --> {{{x^2+x/2+1/16=55/2+1/16}}} --> {{{(x+1/4)^2=440/16+1/16}}} --> {{{(x+1/4)^2=441/16}}}
{{{441/16=21^2/4^2=(21/4)^2}}} or {{{sqrt(441/16)=sqrt(441)/sqrt(16)=21/4}}} so
either {{{x+1/4=21/4}}} --> {{{x=21/4-1/4}}} --> {{{x=20/4}}} --> {{{x=5}}}
or {{{x+1/4=-21/4}}} --> {{{x=-21/4-1/4}}} --> {{{x=-22/4}}} --> {{{x=-11/2}}}
Same solutions to the equation, and the only solution to the geometry problem is {{{x=5}}}.
 
THE QUADRATIC FORMULA
is a formula that derives from completing the square.
I never set to memorize it, but I have been using it for so long that I remember it.
For an equation of the form
{{{ax^2+bx+c=0}}} the solutions are given by the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In the case of {{{2x^2 + x -55 = 0 }}} {{{a=2}}} , {{{b=1}}} and {{{c=-55}}} so
{{{x = (-1 +- sqrt(1^2-4*2*(-55) ))/(2*2) }}}
{{{x = (-1 +- sqrt(1+440))/4 }}}
{{{x = (-1 +- sqrt(441))/4 }}}
{{{x = (-1 +- 21)/4 }}}
So the solutions of the equation are
{{{x=(-1 + 21)/4 }}} --> {{{x=20/4}}} --> {{{x=5}} and
{{{x=(-1 - 21)/4 }}} --> {{{x=-22/4}}} --> {{{x=-11/2}}}
Since {{{_11/2}}} cannot be the with of a rectangle, the only solution is {{{x=5}}} .