Solving
Quadratic Formula
What you need is called the quadratic formula. It is used to solve quadratic equations, which are of the form $ax^2+bx+c=0$. This could be $x^2+3x+2$, for example. With your equation, what you first need to do get $0$ on one side of the equals sign:
$$x^2=3x+4$$
$$0=-x^2+3x+4$$
Okay, now we clearly have a quadratic equation. A good thing to do when using the quadratic formula is defining $a$, $b$, and $c$ (remember the standard form). So, in this case, you'd have
$$a=-1$$
$$b=3$$
$$c=4$$
Next, we literally just plug the numbers into the formula if doing it by hand. The formula is
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
So, we have $$x=\frac{-3 \pm \sqrt{3^2-4*-1*4}}{2*-1}$$
Which reduces to $$x = \frac{-3 \pm 5}{-2}$$
Now, there are almost always two solutions to the quadratic formula (at least in Algebra I textbook problems), because when graphed, quadratic equations take the form of a parabola:
which crosses the x-axis at two places, and we are solving for $x$.
So, the solution is $$x=-1$$ or $$x=4$$ Honestly, I think this is the easiest way to solve a quadratic (though you should know all of them).
Factoring
Factoring operates off of the zero product property (ZPP). Basically, it just says that in an equation $ax=0$ either $a=0$ or $x=0$. Simple, but powerful. So the purpose of factoring is to find expressions that, when simplified, equal the equation you are trying to factor. So, for example, $x^2 - 8x + 12 = 0$ factors out neatly to $(x-6)(x-2)=0$ which, since this is equal to zero, so we can pull out the ZPP and get $x=6$ or $x=2$.
Now, for a nice method to make this easier (as opposed to plug-and-chug). A good way to do it is saying, okay, $a\cdot c$ is equal to whatever and $b$ is equal to whatever. Then, you find two numbers that multiply to $a\cdot c$ and add to $b$. For example, taking our earlier equation, $x^2-8x+12=0$, we see that $a$ is one, so our number has to multiply to 12 and add to -8 (don't forget the negative signs!) and so we can write out the factors of 12, which are 1, 2, 3, 4, 6, and 12. Now, we can change the sign on these numbers (aka, we can make them -1 or 1, for instance) so keep that in mind. Which pair adds to 8? Well, 12-1=11 and 12+1=13, so that's out. -6+-2=-8...oh, perfect.
Then, you say, okay, this is really equal to $x^2 -6x -2x+12=0$. Then, we put parentheses around it: $(x^2-6x)+(-2x+12)=0$. Then, we factor out the largest common factor from each: $x(x-6)+2(x-6)=0$. Once you are finished factoring you should get the same expression in both sets of parentheses. If not, you have done something wrong. Then, you can say $(x-2)(x-6)=0$. From here, we're home free, and we can solve using the ZPP (see the very first example in this section).
Completing the Square
Okay, completing the square is another method to solve a quadratic. First, let's assume you've already set your equation up so that it is directly equal to $0$. Some quadratics are a bit easier to solve by factoring than others. Take $x^2+6x+9=0$, for instance. You can factor this and get $(x+3)^2=0$, and then take the square root (nothing happens to the zero) and notice that $x=-3$. These types of simple equations are called perfect squares.
A whole method has based off of these wonderfully simple equations. Take, for example, the perfectly hideous equation $x^2 +6x -16=0$. We can actually do something really cool here. Let's move the 16 over to the other side (no, don't look at me weird yet, hold on) to get $x^2+6x=16$. Now, let's say we want to make this a perfect square. To get what number in the place of $c$ will make it a perfect square, simply do $(b/2)^2$. In this case, we get 9. So, add 9 to both sides, getting $x^2+6x+9=25$. Then, we know this reduces to $(x+3)^2=25$. How to do that simply? Take $x$ plus $b/2$ and then get $(x+\frac{b}{2})^2$. Square root both sides, and you get $x+3=\pm 5$. Then, remember the way we did this on the quadratic formula, we can just calculate, getting $x=2$ or $x=-8$. Now obviously not all of them turn out quite this nice, but it is another handy method for the quadratic toolbelt.
Deriving the Quadratic Formula
The reason I went through all that is so I could explain where the quadratic formula comes from. Remember that wonderful standard form, $ax^2+bx+c=0$? Well, if you complete the square on that, you get the quadratic formula. This is not that bad. So, here we go:
$$ax^2+bx+c=0$$ $$-c=ax^2+b$$ $$\frac{-c}{a}=x^2+\frac{b}{a}x$$ Side note: $\frac{b}{a}$ divided by 2 is $\frac{b}{2a}$, that squared is $\frac{b^2}{4a^2}$. Carrying on... $$\frac{b^2}{4a^2}+\frac{-c}{a}=x^2+\frac{b}{a}+\frac{b^2}{4a^2}$$ Another side note: To simplify further, we need to give the two fractions on the left side of the equation a common denominator, so $\frac{-c}{a}$ is multipled by $\frac{4a}{4a}$ to give $\frac{b^2-4ac}{4a^2}$ (once the two fractions are added). $$\frac{b^2-4ac}{4a^2}=x^2+\frac{b}{a}+\frac{b^2}{4a^2}$$ Complete the square here $$\frac{b^2-4ac}{4a^2}= (x+\frac{b}{2a})^2$$ Square root both sides $$\sqrt{\frac{b^2-4ac}{4a^2}}= \sqrt{(x+\frac{b}{2a})^2}$$ $$\frac{\pm \sqrt{b^2-4ac}}{2a}=x+\frac{b}{2a}$$ $$\frac{-b \pm \sqrt{b^2-4ac}}{2a}=x$$
Coding
Now, for the coding solution. I used python:
import math
a = 1
b = 3
c = 2
def quad_solve(a, b, c):
if (b*b >= 4*a*c):
print "There is a solution!"
d = math.sqrt((b*b)-(4*a*c))
solution1 = (-b-math.sqrt(d))/(2*a)
solution2 = (-b+math.sqrt(d))/(2*a)
if (solution1 != solution2):
print (solution1,solution2)
else:
print solution1
else:
print "No solutions, imaginary number"
quad_solve(a, b, c)
To plug in the numbers you want, you change a, b, and c (remember the standard equation). The way it works is this. First, there is a function defined that takes in variables a, b, and c. Then, it checks to make sure that the numbers under the square root don't come out negative (if they did, the solution would be an imaginary number). If it passes that, then it literally calculates using the variables, if it doesn't, it tells you so. The other if statement checks to make sure solution 1 and solution 2 aren't equal before printing them both; that's handy on perfect squares.
Hope this helps! Let me know if you have any questions about deriving the quadratic formula, or one of the methods, or anything like that. Finally, one last tip: I had algebra I last year, and the quadratic formula is really important to memorize. Try singing it to the tune of Pop Goes the Weasel; I promise you, you'll have it stuck in your head forever. =)
