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

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Answer 1 · 2017-09-11T16:34:24

#1/5±(2i)/5sqrt194#

#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.