How do you solve #x^2 – 3x – 23 = 5#?

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Answer 1 · 2016-03-29T17:02:23

The solutions are:
#x= color(blue)(7#

#x = color(blue)(-4#

#x^2 - 3x - 23 = 5#

#x^2 - 3x - 23 - 5 = 0#

#x^2 - 3x - 28 = 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=-3, c=-28#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (-3)^2-(4*1* ( -28))#

# = 9 + 112= 121#

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

#x = (-(-3)+-sqrt(121))/(2*1) = (3+- 11)/2#

#x = (3+ 11)/2 = 14/2 = color(blue)(7#

#x = (3-11)/2 = -8/2 = color(blue)(-4#