How do you solve #2x^2-3x-1=0# using the quadratic formula?

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Answer 1 · 2015-08-10T07:23:10

The solutions for the equation are:

#color(blue)(x=(3+sqrt17)/4)#

#color(blue)(x=(3-sqrt17)/4)#

#2x^2−3x−1=0 #

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=2, b=-3, c=-1#

The Discriminant is given by:
#Delta=b^2-4*a*c#
# = (-3)^2-(4*(2)*-1)#
# = 9+8 = 17#

The solutions are found using the formula

#x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-3)+-sqrt(17))/(2*2) = (3+-sqrt(17))/4#

#color(blue)(x=(3+sqrt17)/4)#

#color(blue)(x=(3-sqrt17)/4)#