How do you solve #x^2 - 4x - 32 = 0#?

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Answer 1 · 2018-03-23T23:22:56

The root of the equation is #x = 2+-6# or #x = 8# and #x = -4#.

For an equation in the form of #ax^2 + bx + c= 0#, the quadratic formula (#x=(-b+-sqrt(b^2-4ac))/(2a)#) can be applied.

Plugging in #1# for #a#, #-4# for #b#, and #-32# for #c#:

#x=(-(-4)+-sqrt((-4)^2-4(1)(-32)))/(2(1))#

#x=(4+-sqrt(16-4(-32)))/2#

#x=(4+-sqrt(16-(-128)))/2#

#x=(4+-sqrt(144))/2#

#x=4/2+-sqrt(144)/2#

#x=2+-12/2#

#x=2 +- 6#

Doing addition:

#x=2+6 = 8#

Doing subtraction:

#x=2-6 = -4#