Often, the simplest way
to solve "ax^{2} + bx + c = 0"
for the value of x is to factor
the quadratic,
set each factor equal to zero, and then solve each factor. But sometimes
the quadratic is too messy, or it doesn't factor at all, or you just don't
feel like factoring. While factoring may not always be successful, the
Quadratic Formula can always find the solution.

The Quadratic Formula uses
the "a",
"b",
and "c"
from "ax^{2} + bx + c",
where "a",
"b",
and "c"
are just numbers; they are the "numerical coefficients" of the
quadratic equation they've given you to solve. The Quadratic Formula is derived from the process of completing the square, and is formally stated as:

For ax^{2} + bx + c = 0,
the value of x is given by:

For the Quadratic Formula
to work, you must have your equation arranged in the form "(quadratic)
= 0".
Also, the "2a"
in the denominator of the Formula is underneath everything above,
not just the square root. And it's a "2a"
under there, not just a plain "2". Make
sure that you are careful not to drop the square root or the "plus/minus"
in the middle of your calculations, or I can guarantee that you will forget
to "put them back" on your test, and you'll mess yourself up. Remember
that "b^{2}"
means "the square of ALL of b,
including its sign", so don't leave b^{2} being negative, even if b is negative, because the square of a negative is a positive.

In other words, don't be
sloppy and don't try to take shortcuts, because it will only hurt you
in the long run. Trust me on this!

Here are some examples
of how the Quadratic Formula works:

Solve x^{2} + 3x – 4 = 0

This quadratic happens
to factor:

x^{2} + 3x – 4 = (x + 4)(x – 1) = 0

...so I already know
that the solutions are x = –4 and x = 1. How would my
solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution
looks like this:

Then, as expected, the
solution is x = –4, x = 1.

Suppose you have ax^{2} + bx + c = y,
and you are told to plug zero in for y.
The corresponding x-values
are the x-intercepts of the graph. So solving ax^{2} + bx + c = 0 for x means, among other things, that you are trying to find x-intercepts.
Since there were two solutions for x^{2} + 3x – 4 = 0,
there must then be two x-intercepts
on the graph. Graphing, we get the curve below:

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As you can see, the x-intercepts
(the red dots above) match the solutions, crossing the x-axis
at x = –4 and x = 1. This shows the
connection between graphing and solving: When you are solving "(quadratic)
= 0", you are finding the x-intercepts
of the graph. This can be useful if you have a graphing calculator, because
you can use the Quadratic Formula (when necessary) to solve a quadratic,
and then use your graphing calculator to make sure that the displayed x-intercepts
have the same decimal values as do the solutions provided by the Quadratic
Formula.

Solve 2x^{2} – 4x – 3 = 0. Round your answer
to two decimal places, if necessary.

There are no factors
of (2)(–3)
= –6 that add up
to –4,
so I know that this quadratic cannot be factored.
I will apply the Quadratic Formula. In this case, a = 2, b = –4, and c = –3:

Then the answer is x = –0.58, x = 2.58,
rounded to two decimal places.

Warning: The "solution"
or "roots" or "zeroes" of a quadratic are usually
required to be in the "exact" form of the answer. In the example
above, the exact form is the one with the square roots of ten in it. You'll
need to get a calculator approximation in order to graph the x-intercepts
or to simplify the final answer in a word problem. But unless you have
a good reason to think that the answer is supposed to be a rounded answer, always go with the exact form.

Compare the solutions
of 2x^{2} – 4x – 3 = 0 with the x-intercepts
of the graph:

Just as in the previous
example, the x-intercepts
match the zeroes from the Quadratic Formula. This is always true. The "solutions"
of an equation are also the x-intercepts
of the corresponding graph.