How do you solve $3z^2 + z - 1 = 0$ using the quadratic formula?

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Answer 1 · 2018-04-03T12:35:18

$x=(-1+-sqrt13)/6$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

In your equation which is in the general form
$az^2+bz+c=0$

$a=3, b=1, c=-1$

So putting it into the quadratic formula,

$x=(-1+-sqrt(1+12))/6$
$x=(-1+-sqrt13)/6$

How do you solve 3z^2 + z - 1 = 03z^2 + z - 1 = 0 using the quadratic formula?