How do you solve #x^2-4x=6#?

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Answer 1 · 2015-05-24T00:04:58

First, subtract 6 from both sides:

#x^2 - 4x - 6 = 0#

No combination of two factors of -6 will add up to -4, so you'll have to use the quadratic formula to solve this.

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

#x = (-(-4) +- sqrt((-4)^2-4(1)(-6)))/(2(1))#

#x = (4 +- sqrt(16-(-24)))/2#

#x = 2 +- sqrt(4 - (-6))#

#x = 2 +- sqrt10#

Final Answer