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

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Answer 1 · 2015-05-24T00:04:58

First, subtract 6 from both sides:

$x^{2} - 4 x - 6 = 0$ $x^2 - 4x - 6 = 0$

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2-4ac))/(2a)$

$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (- 6)}}{2 (1)}$ $x = (-(-4) +- sqrt((-4)^2-4(1)(-6)))/(2(1))$

$x = \frac{4 \pm \sqrt{16 - (- 24)}}{2}$ $x = (4 +- sqrt(16-(-24)))/2$

$x = 2 \pm \sqrt{4 - (- 6)}$ $x = 2 +- sqrt(4 - (-6))$

$x = 2 \pm \sqrt{10}$ $x = 2 +- sqrt10$

Final Answer

How do you solve x2−4x=6x^2-4x=6?