How do you solve #3x^2 - 2x = 4# using the quadratic formula?

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Answer 1 · 2016-02-27T04:10:16

The solutions are :
#color(blue)(x=(1+sqrt13)/3#

#color(blue)(x=(1-sqrt13)/3#

#3x^2-2x-4=0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=3, b=-2, c=-4#

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

# = (-2)^2-(4*3*-4)#

# = 4 +48 = 52#

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

#x = (-(-2)+-sqrt(52))/(2*3) = (2+-sqrt(52))/6#

Upon further simplification #sqrt52= sqrt(2*2*13)= 2sqrt13#

So, #x = (2+-2sqrt(13))/6 = (cancel2(1+-sqrt13))/cancel6#
#= (1+-sqrt13)/3#

The solutions are :
#color(blue)(x=(1+sqrt13)/3#
#color(blue)(x=(1-sqrt13)/3#