Solve $3 = -x^2+4x$ by factoring I have $3 = -x^2 + 4x$ and I need to solve it by factoring. According to wolframalpha the solution is $x_1 = 1, x_2 = 3$.
\begin{align*}
 3 & = -x^2 + 4x\\
 x^2-4x+3 & = 0
\end{align*}
According to wolframalpha $(x-3) (x-1) = 0$ is the equation factored, which allows me to solve it, but how do I get to this step? 
 A: You moved everything onto the same side of the equation, which is a great start!
The next step is to understand factorization of quadratic polynomials.
Suppose you have a polynomial of the form
$$x^2 + cx + d \tag{1} $$
You want something of the form
$$ (x+a)(x+b) $$
When you expand out $(x+a) (x+b)$, you get
$$ (x+a)(x+b) = x^2 + (a+b)x + ab \tag{2}$$
Now, when we compare the coefficients of (2) to (1), we see that
$$ a+b = c $$
$$ ab = d $$
That is, we need to find two numbers $a$ and $b$ that add up to $c$ and multiply into $d$.  
In our case, $c$ is -4 and $d$ is 3. Now we have to think about it for a bit and do a bit of guessing and checking, but you should be able to see that $a=-3$ and $b =-1$ meets this criteria.  
-3+(-1) = -4, and (-3)*(-1)=3.
A: Move everything to one side then factor.  Factoring something like this is just trial and error.  You need to find two numbers whose product is $3$ and whose sum is $-4$.  So poking around a bit you should come up with $-3$ and $-1$.  So it factors as:
$x^2-4x+3=(x-3)(x-1)$
A: Here's a similar problem. Let's solve $$x^2 + 6 = 7x$$ by factoring.
First, we need to write this in standard form: $$x^2 - 7x + 6 = 0$$
Then we seek two numbers whose sum is $-7$, and whose product is $6$. The two such numbers are $-1$ and $-6$. 
At this point, you could just jump straight to $(x - 1)(x-6) = 0$, but if you have never seen this before, it might help to rewrite the middle term and factor by grouping. To wit: 
$$x^2 - 7x + 6 = x^2 -1x - 6x + 6$$
We then write this as $$x(x-1) -6(x-1) = (x-6)(x-1)$$
From here, we set each factor equal to zero. 
A: Solving by factoring means reducing the polynomial to a lower degree by dividing it by factors $(x-x_i)$ for already known roots $x_i$.
The reason for this is that the lower degree polynomial might be easier to solve. Here this is not necessary, as a order $2$ polynomial is already solvable with moderate effort.
How to know roots? For this kind of task one should always try the $0, \pm 1, \pm 2$, which here gives $x_1 = 1$. 
What is left is the division:
\begin{align}
& x^2 - 4x + 3 : x - 1 = x - 3 \\
-&(x^2-x) \\
= & -3x+3 = -3(x-1)
\end{align}
So we have the factored polynomial $(x+1)(x-3)$.
A: $$ 3 = -x^2 +4x \implies \ x^2 -4x+3 = 0$$
$$ x^2 - 3x - x +3 = 0$$
$$ x(x-3)-1(x-3)$$
$$ (x-3)(x-1) = 0$$
$$x = 1, x = 3$$
For more on the methods, visit this link.
Alternatively, you could use the quadratic formula:
$$x =  \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
This was obtained from the quadratic equation:
$$ax^2 + bx +c = 0$$
where $a = 1, b = -4, c = 3$
A: $$3=-x^2+4x$$ is equivalent to $$x^2-4x+3=0$$
wich is a quadratic equation, if $x_1$ and $x_2$ are there solutions, we know that $$x^2-4x+3=(x-x_1)(x-x_2)=0$$ where $$x_1+x_2=-(-4)$$
and $$x_1*x_2=3$$
A: \begin{align}x^2-4x+3&=x^2-3x-x+3\\
&=x(x-3)-(x-3)\\
&=(x-1)(x-3)
\end{align}
Note: You could also see that the sum of coefficients is zero, hence one root is $x=1$. Now divide the quadratic by $x-1$ to get the other factor.
