How do you solve #2r^2+r-14=0# using the quadratic formula?

1 Answer
Dec 26, 2016

Answer:

The two solutions are #r = (-1+sqrt(113))/4# and #r = (-1-sqrt(113))/4#

or #r = 2.4075# and #r = -2.9075#

Explanation:

Since this question is given in standard form, meaning that it follows the form: #ax^(2) + bx + c = 0#, or #ar^(2) + br+ c = 0# in this case, 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 #2r^(2)# for this question.#b# is the number that has the #x# variable associated with it and it would be #1r#, and #c# is a number by itself and in this case it is -14.

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

#r = (- (1) +- sqrt((1)^(2) - 4(2)(-14)))/(2(2))#

#r = (-1 +-sqrt(1+112))/4#

#r = (-1 +- sqrt(113))/4#

For these type of problems, you will obtain two solutions because of the #+-# part. So what you want to do is add -1 to #sqrt(113)# together and divide that by 4:

#r = (-1+sqrt(113))/4#
#r = 9.6301/4= 2.4075#

Now, we subtract #sqrt(113)# from -1 and divide by 4:

#r = (-1-sqrt(113))/4#
# r = -11.6301/4 = -2.9075#

Therefore, the two possible solutions are:
#r = 2.4075# and #r = -2.9075#