How do you solve #x^2 + 4x + 4 = 7#?

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Answer 1 · 2015-07-27T16:28:58

The solutions are:
#color(green)(x=-2+sqrt(7)#
#color(green)( x=-2-sqrt(7)#

#x^2+4x+4=7#

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

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=4, c=-3#

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

# = (4)^2-(4*1*((-3))#

# = 16+12 =28#

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

#x = ((-4)+-sqrt(28))/(2*1) = (-4+-2sqrt(7))/2#

# =(cancel(2)(-2+-sqrt(7)))/cancel2#

#color(green)(x=-2+sqrt(7), x=-2-sqrt(7)#