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

1 Answer
Dec 26, 2016

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.htmlhttps://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