**Example: Solve the following equation for x**

**0 = x^2 +9x + 20**

We can solve for x using factoring if possible or the quadratic equation if we cannot factor. The quadratic formula will always solve quadratic equations for x.

**This equation will factor (but note that many quadratic equations will not factor):**

(x + 4)(x + 5) = 0

Therefore, we can solve for both values of x

x + 4 = 0

x = -4

**AND**

x + 5 = 0

x = -5

**Solution {-4,-5}**

**We can also use the quadratic formula and will get the same result:**

**(-b +/- sqrt(b^2 – 4ac)) / [2a]**

For x^2 +9x + 20

a = 1, b = 9, c = 20

(-b +/- sqrt(b^2 – 4ac) / [2a]

(-9 +/- sqrt(9^2 – 4(1)(20)) / [2(1)]

(-9 +/- sqrt (81 – 80) )/ [2]

(-9 +/- sqrt (1) )/ [2]

(-9 +/- 1 )/ [2]

There are two answers:

(-9 -1)/2 = -10/2 = -5

AND

(-9 +1)/2 = -8/2 = -4

**This is the same solution we got using factoring and is {-4,-5}**

**To check your answers:**

Check that both solutions for x are correct – plug each back into the original equation.

**Checking on -5**

0 = x^2 +9x + 20

0 = (-5)^2 +9(-5) + 20

0 = 25 – 45 + 20

0 = -20 + 20

0 = 0 YES

So -5 is a solution

**Checking on -4**

0 = x^2 +9x + 20

0 = (-4)^2 +9(-4) + 20

0 = 16 – 36 + 20

0 = -20 + 20

0 = 0 YES

So -4 is a solution