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

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Answer 1 · 2015-07-25T14:03:23

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

The equation #2x^2−x−4# : is of the form #color(blue)(ax^2+bx+c=0# where:

#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#