Question 14442
From your story, it appears that you have learned to solve quadratic equations by factoring. This is only one of a few ways to solve these type of equations.  The difficulty arises when the equation is not readily factorable.

A quadratic equation can always be solved by use of the "quadratic formula".  While it may look a little formidable, its derivation is not difficult, and it is quite easy to apply if you remember to put you quadratic equation into the "standard form: {{{ax^2 + bx + c = 0}}}

The quadratic formula is:  {{{x = (-b+-sqrt(b^2 - 4ac))/2a}}}

Let's solve an example quadratic equation using this formula:

{{{x^2 + 2x - 8 = 0}}}  This already in the standard form, so: a = 1, b = 2, and c = -8  Substitute these values into their corresponding places in the formula.

{{{x = (-2+-sqrt(2^2 - 4(1)(-8)))/2(1)}}} Now perform the indicated arithmetic to solve for x.

{{{x = (-2+-sqrt(4+32))/2}}}
{{{x = (-2+-sqrt(36))/2}}}
{{{x = (-2/2)+(6/2)}}} and/or {{{x = (-2/2)-(6/2)}}}
{{{x = -1 + 3}}} and/or {{{x = -1 - 3}}}
{{{x = 2}}} and/or {{{x = -4}}}

As a check, we can also solve this equation by factoring:

{{{x^2 + 2x - 8 = 0}}} Factor.
{{{(x - 2)(x + 4) = 0}}} Apply the zero products principle.
{{{(x - 2) = 0}}} and/or {{{(x + 4) = 0}}}

If {{{x - 2 = 0}}}, then {{{x = 2}}}
If {{{x + 4 = 0}}}, then {{{x = -4}}}

So, {{{x = 2}}} and/or {{{x = -4}}}