How do you solve $x^2 -x = 5x - 9$ ?

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Answer 1 · 2015-07-10T09:03:43

$x=3$

$x^2-x=5x-9$

Subtract $5x$ from both sides of the equation.

$x^2-x-5x=-9$ =

$x^2-6x=-9$

Add $9$ to both sides.

$x^2-6x+9=0$

We now have a quadratic equation with the form $ax^2+bx+c=0$ , where $a=1;$ $b=-6;$ and $c=9$ .

We can solve for $x$ using the quadratic formula.

Quadratic Formula

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

$x=(-(-6)+-sqrt(-6^2-4*1*9))/(2*1)$ =

$x=(6+-sqrt(36-36))/2$ =

$x=(6+-sqrt0)/2$ =

$x=(6+-0)/2$ =

$x=6/2$ =

$x=3$

How do you solve x^2 -x = 5x - 9x^2 -x = 5x - 9?