How do you solve #x^2+3x+2=0# using the quadratic formula?

1 Answer
Dec 26, 2016

Answer:

The two solutions are #x = -1 # and #x = -2#

Explanation:

Since this question is given in standard form, meaning that it follows the form: #ax^(2) + bx + c = 0#, we can use the quadratic formula to solve for x:

http://www.1728.org/quadratc.htm
I think it's worthwhile to mention that #a# is the number that has the #x^2# term associated with it. Thus, it would be #1x^(2)# for this question. #b# is the number that has the #x# variable associated with it and it would be #3x#, and #c# is a number by itself and in this case it is 2.

Now, we just plug our values into the equation like this:

#x = (- (3) +- sqrt((3)^(2) - 4(1)(2)))/(2(1))#

#x = (-3 +-sqrt(9-8))/2#

#x = (-3 +- sqrt(1))/2#

For these type of problems, you will obtain two solutions because of the #+-# part. So what you want to do is add -3 and #sqrt(1)# together and divide that by 2:

#x = (-3+sqrt(1))/2#
#x = -2/2= -1#

Now, we subtract #sqrt(1)# from -3 and divide that by 2:

#x = (-3-sqrt(1))/2#
# x = -4/2 = -2#

Therefore, the two possible solutions are:
#x = -1 # and #x = -2#