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How do you solve equation with quadratic formula #3n^2-4n-15=0#?

1 Answer

Nov 1, 2016

#n_1=-5/3; n_2=3#

Explanation:

Since b-4 is even, you can use the formula:

#x=(-b/2+-sqrt((b/2)^2-ac))/a#, so you have

#n=(-(-4/2)+-sqrt((4/2)^2-3*(-15)))/3#
#=(2+-sqrt(4+45))/3#
#=(2+-7)/3#
#n_1=-5/3; n_2=3#

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