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

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Answer 1 · 2016-11-01T06:07:31

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

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

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

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

How do you solve equation with quadratic formula 3n^2-4n-15=03n^2-4n-15=0?