How do you solve #4x^2=12x-11#?

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Answer 1 · 2015-09-22T16:43:13

The solutions are
#color(blue)( (12+sqrt(-32))/8#

#color(blue)( (12-sqrt(-32))/8#

#4x^2=12x−11#

#4x^2 - 12x + 11 = 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=4, b=-12, c=11#

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

# = (-12)^2-(4* 4 * 11)#

# = 144-176 = -32#

As #Delta<0# there are no real solutions

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

#x = (-(-12)+-sqrt(-32))/(2*4) = (12+-sqrt(-32))/8#

The solutions are
#color(blue)( (12+sqrt(-32))/8#
#color(blue)( (12-sqrt(-32))/8#