how to solve $x^2 = 3x + 4$ I am a programmer in the eighth grade taking algebra 1. I am only about a month into the school year, and I need to know how to solve something similar to this equation:
$x^2 = 3x + 4$
However, the problem is that whenever I try to get the square root of both sides of the equation to get rid of the $x^2$, I get something like this:
$x = \sqrt{3x + 4}$
After this, I couldn't figure out what to do, because my only option was to square both sides of the equation, which would get me back where I started. So, how would I solve this? Please make the answer simple enough so that an algebra 1 student would understand. Also, sorry for the tag that probably isn't  great, my rep is so low I cant create a new tag (I would probably tag this as algebra 1).
 A: There are several different techniques to solve this particular equation, which is called a quadratic equation.  Here are two methods that you will learn in Algebra 1.
Method 1:  Factoring
The equation $$x^2=3x+4$$ can be rewritten as $$x^2 -3x -4=0$$
Now the goal is to factor the left-hand side of the equation -- that is, to write it in the form
$$(x + \textrm{something})(x + \textrm{something else}) = 0$$
The "something" and "something else" are to be two numbers whose product is $-4$ and whose sum is $-3$.  Since the product is negative, one of the numbers must be positive and the other one is negative.  After a little playing around, we find that
$$(x+1)(x-4)=0$$ works.  (To check this, multiply each term in the first pair of parentheses times each term in the second pair of parentheses, getting four terms; be careful with negative signs, and combine like terms.)
Finally, we use the zero-product property of the real numbers:  if the product of two quantities is zero, one of the two quantities must be zero.  In this case, the product of $x+1$ and $x-4$ is zero, so either $x+1=0$ (in which case $x=-1$) or $x-4=0$ (in which case $x=4$).  We conclude that there are two solutions to the equation:  $x=-1$ or $x=4$.  You can check that both of these work in the original equation.
Method 2:  Completing the square
Rewrite the equation as
$$x^2 - 3x = 4$$
Now, look at the coefficient of the $x$ term, divide it by $2$, and square the result.  Add this to both sides of the equation.  In our case, the coefficient of the $x$ term is $-3$, half of which is $-3/2$, and the square of that is $9/4$, so we add $9/4$ to both sides of the equation:
$$x^2 - 3x + \frac{9}{4} = 4 + \frac{9}{4} = \frac{25}{4}$$
Now, why did we do this?  We did this because the left-hand side of the equation can now be recognized as the square of $(x - \frac{3}{2})$.  That is, we have
$$\left(x-\frac{3}{2}\right)^2 = \frac{25}{4}$$
Taking the square root of both sides, we end up with
$$x - \frac{3}{2} = \pm \frac{5}{2}$$
so
$$x = \frac{3}{2} \pm \frac{5}{2}$$
Reading this equation with the plus sign, we have $x = 8/2 = 4$; reading it with the minus sign, we have $x = -2/2 = -1$.
A: The standard method is to multiply everything by $4$ and transfer the $x$ term on the left-hand side:
$$
4x^2-12x=16
$$
Now recall $(a+b)^2=a^2+2ab+b^2$ and observe that we can take $a=2x$, so from $2ab=-12x$ we obtain $4bx=-12x$, that's satisfied for $b=-3$. We need the $b^2$ term, so we add it on both sides:
$$
4x^2-12x+9=16+9
$$
The left-hand side can be rewritten $(2x-3)^2$, so we are reduced to
$$
(2x-3)^2=25
$$
that gives us
$$
2x-3=5\qquad\text{or}\qquad 2x-3=-5
$$
A different strategy is to set $x=t+a$ and try to determine $a$ in such a way that some term vanishes:
$$
t^2+2at+a^2=3t+3a+4
$$
If we set $2a=3$, the $t$-term disappears:
$$
t^2+\frac{9}{4}=\frac{9}{2}+4
$$
and so
$$
t^2=\frac{25}{4}
$$
and therefore $t=5/2$ or $t=-5/2$. Thus
$$
x=\frac{5}{2}+\frac{3}{2}
\qquad\text{or}\qquad
x=-\frac{5}{2}+\frac{3}{2}
$$
You'll learn the “quadratic formula” that can be obtained in the same way as the first method above: if $ax^2+bx+c=0$, then
$$
x=\frac{-b+\sqrt{b^2-4ac}}{2a}
\qquad\text{or}\qquad
x=\frac{-b-\sqrt{b^2-4ac}}{2a}
$$
provided $b^2-4ac\ge0$. In your case $a=1$, $b=-3$ and $c=-4$.
A: What you need to get rid of is the term proportional to $x$. How to do this can be seen by looking at the binomial formula
$$(x+c)^2 = x^2 + 2cx + c^2$$
You see, on the left hand side there's a pure square, while on the right hand side, there's a term having an $x$. So what you need to figure out is the correct $c$ to use.
A: Have you learned the distributive law of multplication yet
$3(4+3) = 3\cdot 4 + 3 \cdot3$
If we had an $x$ in there instead of a 4...
$3(x+3) = 3x + 3 \cdot3$
Lets take this up one more degree of difficulty.
$(x+1)(x+3) = (x+1)x + (x+3)3 = x^2 +x + 3x + 9 = x^2 + 4x + 9$
We have to do the same thing in reverse.  And, it takes some guesswork and intuition.
$x^2 - 3x - 4 = 0$
We want to find factors $(x+a)(x+b) = 0$ for this such that
$ab = -4$
and
$a+b = -3$
If the solutions are rational (and they are not always)
then $a,b$ need to be factors of $4$
that gives us $4,1$ and $2,2$ to work with, and it turns out one of them works
$x^2 - 3x - 4 = (x-4)(x+1)= 0$
Now, if $(x-4)(x+1)= 0$ then
If $(x-4) = 0$ or if $(x+1)= 0$ then the whole thing equals 0.
$x = 4$ or $x = -1$
