How to find $x^2 - x$? I'm quite a novice when it comes to maths. I'm on a problem in which I have had to isolate $x$ , through factorials which I completed without problem. However, now I am stuck on a seemingly more minor problem.
The problem I currently have is $x^2 - x = 380$. I know that this can be solved for $x = 20$, however I am unsure how this has been worked out.
I am sorry for this being such a basic question, however I simply have no idea how this was solved.
Thanks,
 A: 
Quadratic Formula:  We consider the equation $ax^2+bx+c=0$ where $a,b,c\in\mathbb R$ and $a\neq0$, then its solutions are given by the formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
  where we have 3 cases:
  
  
*
  
*If $b^2-4ac\gt0$ then the equation have two solutions.
  
*If $b^2-4ac=0$ then the equation have one and unique solution.
  
*If $b^2-4ac\lt0$ then the equation have no real solutions.
  

Now for your particular problem you have $a=1$, $b=-1$ and $c=-380$. Can you find the solution based on that formula?
A: You may recognize that $x^2-x=x(x-1)$ and that $380=20\cdot 19$, which will give you $x=20$.
Writing $380=(-20)\cdot (-19)$ will give you $x=-19$.
A: This is a quadratic polynomial, this can always be solved using the quadratic formula as was mentioned in a comment, or sometimes by factoring the quadratic. Here we can factor like so:
$0 = x^2 - x - 380 = (x+19)(x-20)$
Now on the right hand side we simply have the product of two numbers, this can only equal zero if one of the two is equal to zero. This gives us:
$(x + 19) = 0$ or $(x-20) = 0$    
or equivalently: 
$x = -19$ or $x = 20$
Those are the two solutions.
A: At the moment, you have $x^2-x=380$, which is the same as $x^2-x-380=0$. In general, a function with degree two can be written as $ax^2+bx+c=0$, so in this case, $a=1$, $b=-1$ and $c=-380$.
Now you can compute the discriminant $D$ as $D=b^2-4ac=(-1)^2-4*1*(-380) = 1 + 4*380 = 1521$.
Because the value of D is strictly positive, there are two roots. To find those roots, you can use the following formula:
$x_1=\frac{-b-\sqrt{D}}{2a}= \frac{-(-1)-\sqrt{1521}}{2}=\frac{1-39}{2}=\frac{-38}{2}=-19 \\$
$x_2=\frac{-b+\sqrt{D}}{2a} = \frac{-(-1)+\sqrt{1521}}{2}=\frac{1+39}{2}=20\\$ 
So the two solutions are $-19$ and $20$.
A: $$x^2 - x = 380$$
Rewrite this as:
$$x^2-x-380=0$$
Use quadratic formula to solve it now. It is:
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
For:
$$ax^2+bx+c=0$$
Just plug this in:
$$x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-380)}}{2(1)}=\frac{1 \pm \sqrt{1521}}{2}=\frac{1 \pm 39}{2}$$
Which means:
$$\therefore x=20 \text{ or } -19$$
A: A somewhat different way of putting this is that if you know that $x=20$ is a solution to $x^2-x=380$ then that implies that you can write
$$x^2-x-380 = 0 \Leftrightarrow (x-20) f(x) =0,$$
for some function $f$.  In this case $f(x)=x+19$ so the other value of $x$ that makes $x^2-x-380$ equal to zero is $x=-19$.
A: This particular problem offers another solution angle:  factorization.  We have
$$380=x^2-x=x(x-1)$$
So now we can see that $380$ is a number multiplied by that number less one.  $380$ is between $400=20^2$ and $361=19^2$, so it is reasonable to guess that $x=20$ and it is very easy to see that $20\cdot19=380$, and we are done.
On a separate note, we can also take note that $x^2-x$ is a special form and can be rewritten more compactly in "binomial form" as
$$x^2-x=2{x\choose 2}=380$$
In this form, the binomial series in the quadratic line is the only term, so we can look directly to this series for the answer, which is $x=20$.
