We start with a definition.
A quadratic equation in x is an equation of the form
a x^2 + b x + c = 0
where a, b, c are real numbers and a is not equal to 0.
The quadratic formula allows us to solve any quadratic equation.
However for some quadratic equations,
other methods (like factoring) may be
easier.
Here is the quadratic formula.
Let a, b and c be real numbers with a no equal to 0.
Then the solutions of the equation  a x^2 + b x + c = 0
are given by x = ( -b +/- root(b^2 - 4ac))/(2a)
The notation +/- means that there are two solutions:
one solution corresponds to the plus sign
and another solution corresponds to the
minus sign.
Let's illustrate this with the examples on page 58.
Solve the following quadratic equations:
Question 1:  2x^2 = 5 + 3x
The first thing we have to do is rewrite the equation in standard form ax^2 + bx + c = 0,
So we need to move the 5 and the 3x to the left.
We get : 2x^2 - 3x - 5 = 0
Now we can just read of what a, b and c are:
a = 2 , b = -3 and c = -5.
Using the quadratic formula, we get that
x = ( -b +/- root(b^2 - 4ac))/(2a)
We substitute 2 for a, -3 for b and -5 for c.
And now we simplify.  First, I typically
simplify the radicand.
First, I figure out what b^2 is. In this (-3)^2 = 9.
And then I figure out if if -4ac is positive or negative.
So here we have (-4)*2*(-5) which will be positive.
So I first simplify into the following form.
Now I evaluate how much the radical is.
9 + 4*2*5 = 9 + 40 = 49
So we get  x = (3 +/- root(49))/4.
We can simplify root(49): it is just 7.
Which means that there are two solutions.
One solution corresponds to the minus
sign: x = (3 - 7)/4 which is -4/4
which simplifies to -1.
The other solution corresponds to the plus sign: x= (3 + 7)/4, which is 10/4
which simplifies to 5/2.
So we have two solutions: x = -1 or x = 5/2.
Question 2: Solve the equation 3t^2 + t - 6 = 0.
Here we can immediately use to quadratic formula:
a = 3 , b = 1 and c = -6
So t = ( -b +/- root(b^2 - 4ac))/(2a)
We substitute 3 for a, 1 for b and -6 for c.
We simplify the radical: 1^2 = 1 , -4*3*(-6) is positive.
Now we evaluate the radical: 1 + 4*3*6 = 1 + 72 = 73.
So this means that there are two
solutions again:
one corresponding to the negative sign:
 t=( -1 - root(73))/6
and one corresponding to the plus sign:
 t=( -1 + root(73))/6
Since we can not simplify these radicals any further
we typically use the +/- notation.
So we would write our answer as
 t=( -1 +/- root(73))/6
Question 3: 2x^2 - 8x + 5 = 0
We can immediately use to quadratic formula:
a = 2 , b = -8 and c = 5
So x = ( -b +/- root(b^2 - 4ac))/(2a)
We substitute 2 for a, -8 for b and 5 for c.
Next, we simplified the radicand: (-8)^2 = 64 , -4*2*5 is negative
and so we get 64 -40 for the radicand.
Simplifying, we get ( 8 +/- root(24))/4.
A very common mistake is now to simplify this by dividing 8 by 4.
We cannot do this since the
numerator is not in factored form.
Instead we need to simplify the radical and factor the numerator (if possible).
So root(24): we can write 24 as 4*6
Use the product rule to write the square root as the product of two square roots
and simplify: root(4) = 2.
So x = (8 +/- root(24))/4
We substitute 2 * root(6) for root(24)
and now we can factor the numerator. We have two factors: 8 and 2 * root(6).
We can factor out the greatest common factor, which is 2.
and are left with 4 +/- root(6).
Now, since the numerator is factored, we can simplify the 2 in the numerator and the 4 in the denominator.
and we get 1 and 2.
So our answer becomes: (4 +/- root(6))/4.
And that is our final answer.
Question 4: y^2 - 3y + 4 = 0
Again we can immediately use the quadratic formula.
a = 1 , b = -3 and c = 4
So y = ( -b +/- root(b^2 - 4ac))/(2a).
We substitute 1 for a, -3 for b and 4 for c.
Next we simplify the radicand: (-3)^2 = 9
-4*1*4 will be negative, it is -16.
So we get (3 +/- root(9-16))/2
Simplifying the radical, we get (3 +/- root(-7))/2.
Since we cannot take the square root out of a negative number,
we get that there are no real solutions.
It is true that if you know complex numbers, we could write the solutions as complex numbers.
But here we will only consider real numbers.
So for us this equation will have no solutions.
