How do you solve #2x^2+11x-6=0# using the quadratic formula?

1 Answer
Dec 26, 2016

Answer:

The two solutions are #x = 0.50# and #x = -6#

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:
https://mathbitsnotebook.com/Algebra1/Quadratics/QDquadform.html

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 #2x^(2)# for this question.#b# is the number that has the #x# variable associated with it and it would be #11x#, and #c# is a number by itself and in this case it is -6.

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

#x = (- (11) +- sqrt((11)^(2) - 4(2)(-6)))/(2(2))#

#x = (-11 +-sqrt(121+48))/4#

#x = (-11 +- sqrt(169))/4#

For these type of problems, you will obtain two solutions because of the #+-# part. So what you want to do is add -11 to #sqrt(169)# together and divide that by 4:

#x = (-11+sqrt(169))/4#
#x = 2/4= 0.50#

Now, we subtract #sqrt(169)# from -11 and divide by 4:

#x = (-11-sqrt(169))/4#
# x = -24/4 = -6#

Therefore, the two possible solutions are:
#x = 0.5# and #x = -6#