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

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Answer 1 · 2016-12-11T02:50:05

$x=1+sqrt7,$ $1-sqrt7$

$x^2-2x=6$

Subtract $6$ from both sides.

$x^2-2x-6=0$ is a quadratic equation, $ax^2+bx+c$ , where $a=1$ , $b=-2$ , and $c=-6$ .

If by evaluate you mean solve, use the quadratic formula.

Quadratic Formula

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

Substitute the known values into the formula and solve.

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

$x=(2+-sqrt(4+24))/2$

$x=(2+-sqrt(28))/2$

$x=(2+-sqrt(2xx2xx7))/2$

$x=(2+-2sqrt7)/2$

$x=(cancel(2)^1+-sqrt7)/cancel(2)^1$

$x=1+sqrt7,$ $1-sqrt7$

How do you evaluate x^{2}-2x=6x^{2}-2x=6?