# What kind of solutions does 3z^2 + z - 1 = 0  have?

Jul 5, 2015

The discriminant (the thing we take the square root of in the quadratic formula) is:

${b}^{2} - 4 a c$.

In $3 {z}^{2} + z - 1 = 0$, we have

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

So
${b}^{2} - 4 a c = {\left(1\right)}^{2} - 4 \left(3\right) \left(- 1\right) = 1 + 12 = 13$.

$13$ is positive, so there are two distinct real solutions. It is not a perfect square, so the solutions are irrational.

The equation has two distinct irrational real solutions.