How do you solve #x^2-4x +12=0 #?

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Answer 1 · 2017-06-19T17:10:29

See a solution process below:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Substituting #1# for #a#; #-4# for #b# and #12# for #c# gives:

#x = (-(-4) +- sqrt((-4)^2 - (4 * 1 * 12)))/(2 * 1)#

#x = (4 +- sqrt(16 - 48))/2#

#x = 4/2 +- sqrt(16 - 48)/2#

#x = 2 +- sqrt(16 - 48)/2#

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

#x = 2 +- sqrt(16 * -2)/2#

#x = 2 +- (sqrt(16) * sqrt(-2))/2#

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

#x = 2 +- 2sqrt(-2)#

Or

#2(1 + sqrt(-2))#