How do you solve $2 x^{2} - x - 4 = 0$ $2x^2-x-4=0$ using the quadratic formula?

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Answer 1 · 2015-07-25T14:03:23

The solutions are:
$x = \frac{1 + \sqrt{33}}{4}, x = \frac{1 - \sqrt{33}}{4}$ $color(blue)(x=(1+sqrt(33))/4 , x=(1-sqrt(33))/4$

The equation $2 x^{2} - x - 4$ $2x^2−x−4$ : is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:

$a = 2, b = - 1, c = - 4$ $a=2, b=-1, c=-4$

The Discriminant is given by:
$Delta=b^2-4*a*c$

$= (-1)^2 - (4)*(2)(-4)$

$=1+32$
$=33$

As $Delta>0$ there are two solutions,

The solutions are found using the formula:
$x=(-b+-sqrtDelta)/(2*a)$

$x = (-(-1)+-sqrt(33))/(2*2) = (1+-sqrt(33))/4$
The solutions are:
$color(blue)(x=(1+sqrt(33))/4 , x=(1-sqrt(33))/4$

How do you solve 2x^2-x-4=02x2−x−4=02x^2-x-4=0 using the quadratic formula?