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

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