Question 1014361
{{{3x^2-31x-60=0}}} is a quadratic equation.
That means that it can be written (in this case, it is written)
as a degree 2 polynomial equal to zero.
In simpler words, {{{x}}} appears squared as {{{x^2}}} in the equation;
it may also appear as {{{x}}} , maybe multiplied by a number,
and there is no other strange business with {{{x)))
(no {{{sqrt(x)}}} , no {{{x}}} in a denominator, no {{{sin(x)}}} , no {{{2^x}}} , etc).
 
There are 3 possible methods to solve a quadratic equation:
factoring (which only works if there are rational solutions),
"completing the square", and
using the quadratic formula.
 
In this case, I would use the quadratic formula.
The quadratic formula says that the solutions to an equation of the form
{{{ax^2+bx+c=0}}}, if they exist in the math class you are taking,
are given by {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
For {{{3x^2-31x-60=0}}}<-->{{{3x^2+(-31)x+(-60)=0}}} ,
{{{a=3}}} , {{{b=-31}}} , and {{{c=-60}}} , so
{{{x=(-(-31) +- sqrt((-31)^2-4*3*(-60)))/(2*3) }}}
{{{x=(31 +- sqrt(31^2+4*3*60))/6}}}
{{{x=(31 +- sqrt(961+720))/6}}}
{{{x=(31 +- sqrt(1681))/6}}}
That gives you two solutions that exist,
because the square root of a positive number exists for most math class levels.
In some cases those square roots are irrational numbers, such as {{{sqrt(2)}}} ,
but in this case {{{sqrt(1681)=41}}} turned out to be a nice whole number.
So, {{{x=(31 +- sqrt(1681))/6}}}<-->{{{x=(31 +- 41)/6}}} gives us two solutions
{{{system(x=(31+41)/6=72/6=12,"and",x=(31-41)/6)=(-10)/6=-5/3}}} .
 
THE FACTORING WAY TO SOLVE IT
consists of figuring out that {{{3x^2-31x-60=(x-12)(3x+5)}}}
and re-writing {{{3x^2-31x-60=0}}} as {{{(x-12)(3x+5)=0}}} .
If {{{(x-12)(3x+5)=0}}} , one of those two bracketed factors must be zero, so
{{{system(x-12=0,"or",3x+5=0)}}}--->{{{system(x=12,"or",3x=-5)}}}--->{{{system(x-12=0,"or",x=-5/3)}}} .
Figuring out that {{{3x^2-31x-60=(x-12)(3x+5)}}} requires some thinking,
but you do not have to mess with the quadratic formula.
To begin, you multiply the coefficient of {{{x^2}}} and the independent term (the one without {{{x}}} );
{{{3*(-60)=-180}}} .
Then you look for pairs of factors of that number
that add up to the coefficient of {{{x}}} in your equation: {{{-31}}} .
You look for pairs of factors of {{{180}}} , and worry about the minus sign later.
Sometimes there is a lot of pairs, and you have to organize.
I like to look for the smaller factor starting with {{{1}}} ,
and then see if I can make factor pairs where the smaller factor is  {{{2}}} , {{{3}}}, {{{4}}} , etc, in that order.
{{{180=1*180}}}
{{{180=2*90}}}
{{{180=3*60}}}
{{{180=4*45}}}
{{{180=5*36}}}
{{{180=6*30}}}
{{{7}}} and {{{8}}} are not factors of {{{180}}}
{{{180=9*20}}}
{{{180=10*18}}}
{{{11}}} is not a factor of 180
{{{180=12*15}}}
{{{13}}} and {{{14}}} are not factors of {{{180}}}
and factors equal to {{{15}}} and larger have already been found.
From the list above, we know that
{{{(-4)*45=-180}}} and {{{4*(-45)=180}}} are factor pairs that multiply to yield {{{-180}}} ,
but you want them to add up to {{{-31}}} .
The sum {{{-4+45=41}}} has the wrong sign.
{{{4+(-45)=-41}}} has the expected negative sign, and is close to {{{-31}}} .
Obviously we want to give the minus sign to the larger factor, to make the sum negative.
You also see that {{{4}}} and {{{45}}} are a little to far apart,
and that makes the difference {{{45-4=41}}} larger than {{{31}}}.
So we try a pair of factors that are a little closer together:
{{{5(-36)=-180}}} and {{{5+(-36)=-31}}} ,
and discover that we have found our pair of factors.
Now, you re-write {{{3x^2-31x-60}}} using those factors, like this:
{{{3x^2-31x-60=3x^2+5x-36x+60}}} .
Finally, find common factors, and "take out the common factors", twice:
{{{3x^2+5x-36x+60=(3x^2-36x)+(5x+60)=3x(x-12)+5(x-12)=(3x+5)(x-12)}}}