# How do you find the discriminant and how many and what type of solutions does 2x^2 - 3x + 4 = 0 have?

May 17, 2015

$2 {x}^{2} - 3 x + 4$ is of the form $a {x}^{2} + b x + c$, with $a = 2$, $b = - 3$ and $c = 4$.

The discriminant can be calculated using the formula as follows:

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - \left(4 \times 2 \times 4\right) = 9 - 32 = - 23$

This is negative, so $2 {x}^{2} - 3 x + 4 = 0$ has two distinct complex roots and no real roots.

The possible cases are:

$\Delta > 0$ : Two distinct real roots.
$\Delta = 0$ : One repeated real root.
$\Delta < 0$ : No real roots (two distinct complex ones).