How do you solve #-5m ^ { 2} + 2m + 39= 0#?

1 Answer
Sep 11, 2017

#1/5±(2i)/5sqrt194#

Explanation:

Let's use the quadratic formula.

#x=(-b±sqrt(b^2-4ac))/(2a)#

a = -5
b=2
c=39

Let's plug the numbers in and solve

#x=(-(2)±sqrt((2)^2-4(5)(39)))/(2*(-5)) ->#

#x=(-2±sqrt(4-780))/-10 ->#

#(-2±sqrt(-776))/-10 ->#

#(-2±4isqrt194)/-10 ->#

#1/5±(2i)/5sqrt194#

You get imaginary roots for this equation. If you have not learned about imaginary roots/numbers, the answer would be that there are no roots.