# How do you solve #x^{2} + 3x + 6< 0#?

##### 1 Answer

There are no Real values of

#x^2+3x+6 < 0#

In other words, the solution set is empty.

#### Explanation:

**Method 1**

Complete the square to find:

#x^2+3x+6 = (x^2+3x+9/4)-9/4+6#

#color(white)(x^2+3x+6) = (x+3/2)^2+15/4#

Note that for any Real value of

#(x+3/2)^2 >= 0#

and hence:

#x^2+3x+6 = (x+3/2)^2 +15/4 >= 15/4#

So there is no Real value of

#x^2+3x+6 < 0#

**Method 2**

This has discriminant

#Delta = b^2-4ac = 3^2-4(1)(6) = 9-24 = -15#

Since

Then since the coefficient of

#x^2+3x+6 > 0#

for all Real values of

Hence there are no Real solutions to