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

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Answer 1 · 2015-05-17T16:06:42

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

The discriminant can be calculated using the formula as follows:

$Delta = b^2-4ac = (-3)^2-(4xx2xx4) = 9-32 = -23$

This is negative, so $2x^2-3x+4 = 0$ 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).

How do you find the discriminant and how many and what type of solutions does 2x^2 - 3x + 4 = 02x^2 - 3x + 4 = 0 have?