Question 917290
There is a parallelogram with a "base" measuring {{{b}}} inches and a "height" measuring {{{h=b+7}}} inches.
The area of that parallelogram, in square inches, calculated as {{{A=b*h}}} is
{{{A=b(b+7)}}} and we know that is 60 square inches, so our equation is
{{{b(b+7)=60}}} .
That is, indeed, a quadratic equation.
It be written in many equivalent forms, like
{{{b(b+7)=60}}}<--->{{{b^2+7b=60}}}<--->{{{b^2+7b-60=0}}}
and infinitely many other ways.
 
There are many ways to solve the quadratic equation.
One of them if "by factoring", which involves knowing factors of 60.
A fifth grader that realizes that {{{5*12=60}}}
would figure that {{{highlight(b=5)}}} with {{{b+7=5+7=12}}} is a solution.
Your teacher, however,
would expect you to start from the {{{b^2+7b-60=0}}} form of the equation,
factor the left hand side into two factors to get
{{{(b+12)(b-5)=0}}} and find that the solutions of the equation are
{{{b=-12}}}<--->{{{b+12=0}}} . which makes the first factor zero, and
{{{highlight(b=5)}}}<--->{{{b-5=0}}} which makes the second factor zero.
Since the measurement of the base is a positive number, {{{b=-12}}} makes no sense, {{{highlight(b=5)}}} is the only sensible answer.
The length of the base is{{{highlight(5inches)}}} .
 
You could also solve the quadratic equation by "completing the square" or by using the quadratic formula.
Completing the square:
{{{b^2+7b=60}}}
{{{b^2+7b+(7/2)^2=60+(7/2)^2}}}
{{{(b+7/2)^2=60+49/4}}}
{{{(b+7/2)^2=240/40+49/4}}}
{{{(b+7/2)^2=289/4}}}
{{{(b+7/2)^2=(17/2)^2}}}
So,
either {{{b+7/2=17/2}}}-->{{{b=17/2-7/2}}}-->{{{b=10/2}}}-->{{{highlight(b=5)}}} ,
or {{{b+7/2=-17/2}}}-->{{{b=-17/2-7/2}}}-->{{{b=-24/2}}}-->{{{b=-12}}} , which is not a good length for the base of a parallelogram.
Using the quadratic formula, which says that {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} gives the solutions to {{{ax^2+bx+c=0}}} .
Applied to {{{x^2+7x-60=0}}} , {{{a=1}}} {{{b=7}}} and {{{c=-60}}} .
So the solutions to {{{x^2+7x-60=0}}} are given by
{{{x=(-7 +- sqrt(7^2-4*1*(-60) ))/(2*1)=(-7 +- sqrt(49+240))/2=(-7 +- sqrt(289))/2=(-7 +- 17)/2}}} .
So,
either {{{x=(-7 + 17)/2=10/2=5}}} ,
or {{{x=(-7 - 17)/2=(-24)/2=-12}}} .
So, since the solutions to {{{x^2+7x-60=0}}} are {{{x=5}}} and {{{x=-12}}} ,
the solutions to {{{b^2+7b-60=0}}} are {{{b=5}}} and {{{b=-12}}} . (I just change the name of the variable used for the base so that ift would not get confused with the {{{b}}} in {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}and {{{ax^2+bx+c=0}}} ).
 
NOTES:
What is the base and what is the height of a parallelogram (or a triangle) is
"in the eye of the beholder". We think of the base as what we stand something on so that it will stay stable, and the height as the perpendicular measurement to the farthest point from the base. However, for calculation purposes, we use as base whatever makes calculations easier.
Your parallelogram looks like this:
{{{drawing(400,300,-1,19,-2,13,
line(0,0,5,0),line(12.04,12,17.04,12),
line(0,0,12.04,12),line(5,0,17.04,12),locate(2.5,0,5)
)}}} so it looks a bit unstable if you stand it on that "base".