How do you solve #4x^2 – 12x = -9#?

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Answer 1 · 2015-09-14T06:10:14

#color(blue)(x=3/2#

#4x^2–12x+9=0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=4, b=-12, c=9#

The Discriminant is given by:
#color(blue)(Delta=b^2-4*a*c#
# = (-12)^2-(4*4*9)#
# = 144-144=0#

As #Delta=0# there is only one solution.

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

#x = (-(-12)+-sqrt(0))/(2*4) = (12+0)/8 = 12/8 = 3/2#

The solution is
#color(blue)(x=3/2#