Question 727151
How to determine the number of solutions.
Let's look at the quadratic formula:
{{{x = (-b +- sqrt(b^2 - 4ac))/2a}}}
The key is the expression in the square root: {{{b^2 - 4ac}}}. 
In general there are three cases:<ol><li>{{{b^2 -4ac}}} is positive. 
The square root of a positive number is also some positive number. 
So in the numerator of the quadratic formula we will get two values: 
({{{-b}}} + the square root) and ({{{-b}}} - the square root). 
So when {{{b^2 - 4ac > 0}}} we get {{{two}}}{{{ solutions}}}.
</li><li>{{{b^2 -4ac}}} is zero. 
The square root of zero is zero. So in the numerator we get ({{{-b + 0}}}) and ({{{-b - 0}}}). 
But both of these are equal to {{{-b}}}! 
So when {{{b^2 - 4ac = 0}}} we only get {{{one}}}{{{ solution}}}.
</li><li>{{{b^2 -4ac}}} is negative. What is the square root of a negative number? What can we square and get a negative number as an answer? Answer: {{{Nothing}}}. 
You cannot square any Real number and get a negative. So, when {{{b^2 -4ac < 0}}} there are {{{no}}}{{{ solutions}}}.
</li></ol>

example:
1.
{{{x^2 + 8x - 4 = 0}}}... {{{8^2 - 4*1*(-4) = 0}}} which is {{{64 +16 = 80}}}..=> {{{b^2 - 4ac > 0}}}; so, we have {{{two}}}{{{ solutions}}}

2.

{{{x^2 + 4x + 4 = 0}}}... {{{4^2 - 4*1*(4) = 0}}} which is {{{16-16 = 0}}}..=> {{{b^2 - 4ac = 0}}}; so, we have {{{one}}}{{{ solution}}}

3.

{{{x^2 + 4x + 8 = 0}}}... {{{4^2 - 4*1*(8) = 0}}} which is {{{16-32 = -6}}}..=> {{{b^2 - 4ac <0}}}; so, we have {{{no}}}{{{ solution}}}