The quadratic formula and discriminant were introduced in a prior section in the Algebra II curriculum.
Quickly read through this earlier section and do plenty of the online exercises to make sure that you understand the concepts there.
For your convenience, this current section gives an inanutshell review of the quadratic formula, discriminant, and related concepts.
REVIEW
the quadratic formula, the discriminant, related concepts

every quadratic function $\,y = ax^2 + bx + c\,$, $\,a\ne 0\,$, is a parabola. The parabola has different $x$intercepts based on different values of the discriminant: exactly one $x$intercept two different $x$intercepts no $x$intercepts 
There are three types of solution sets possible for a quadratic equation $\,ax^2 + bx + c = 0\,$, $\,a\ne 0\,$.
These solution sets are determined by the discriminant, $\,b^2  4ac\,$: whether it is zero, positive, or negative.
The three situations are summarized in the table below.
SOLUTIONS OF $ax^2 + bx + c = 0\,$, coefficients are real numbers, $a\ne 0$  
value of the discriminant  $b^2  4ac = 0\ $  $b^2  4ac > 0\ $  $b^2  4ac < 0\ $ 
nature of solution set  exactly one real number solution  two different real number solutions 
no real number solutions; a complex conjugate pair 
the solution(s) are:  $$\displaystyle\,x = \frac{b}{2a}$$ 
$\displaystyle\,x = \frac{b + \sqrt{b^2  4ac}}{2a}$ or $\displaystyle\,x = \frac{b  \sqrt{b^2  4ac}}{2a}$ 
$$ \begin{align} x &= \frac{b + i\sqrt{(b^2  4ac)}}{2a}\cr &= \frac{b}{2a} + i\frac{\sqrt{(b^24ac)}}{2a} \end{align} $$ or $$ \begin{align} x &= \frac{b  i\sqrt{(b^2  4ac)}}{2a}\cr &= \frac{b}{2a}  i\frac{\sqrt{(b^24ac)}}{2a} \end{align} $$ 
Note: Since $\,b^2  4ac < 0\,$, we have $\,(b^2  4ac) > 0\,$. Therefore, $\,\frac{\sqrt{(b^24ac)}}{2a} \ne 0\,$, so these two solutions have the same real part and opposite nonzero imaginary parts, hence form a complex conjugate pair. 
The quadratic expression $\ ax^2 + bx + c\ $ ($\,a \ne 0\,$) is called
irreducible if $\ b^2  4ac < 0\ $.
Equivalently, the quadratic expression $\ ax^2 + bx + c\ $ ($\,a \ne 0\,$) is called
irreducible if the equation $\,ax^2 + bx + c = 0\,$ has no real number solutions.
In this case, the trinomial $\,ax^2 + bx + c\,$ cannot be factored using
real numbers.
It can, however, be factored using the complex conjugate pair that is given by the quadratic formula.
This idea is explored in a future section.
Recall the beautiful relationship between the zeroes of a polynomial and its factors:
EXAMPLE:
Find a quadratic function with zeroes $\,2\,$ and $\,3\,$;
the coefficient of the $\,x^2\,$ term should equal $\,5\,$.
SOLUTION:
Since $\,2\,$ is a zero, $\,x  2\,$ is a factor.
Since $3\,$ is a zero, $\,x  (3) = x + 3\,$ is a factor.
Since the coefficient of the $\,x^2\,$ term must be $\,5\,$, we need a constant factor of $\,5\,$.
(Note that this constant factor does not change the zeroes.)
The function $\,f\,$ defined below is the desired quadratic function:
$$\,f(x) = 5(x2)(x+3) = \ldots = 5x^2 + 5x  30\,$$
EXAMPLE:
Find a quadratic function with real coefficients that has exactly one real zero, $x = 4\,$.
The coefficient of the $\,x\,$ term should equal $\,7\,$.
SOLUTION:
Since $\,4\,$ is a zero, $\,x  4\,$ is a factor.
To get a quadratic function with no different zero, we need two factors of $\,x  4\,$.
Thus, every function of the form $\,k(x4)(x4) = \ldots = kx^2  8kx + 16k\,$ satisfies the zero requirement.
For the coefficient of the $\,x\,$ term to equal $\,7\,$, we need $\,8k = 7\,$, so $k = \frac{7}{8}\,$.
The function $\,f\,$ defined below is the desired quadratic function:
$$\,f(x) = \frac{7}{8}(x4)(x4) = \ldots = \frac{7}{8}x^2 + 7x  14\,$$
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
