How do you solve using the quadratic formula $6 x^{2} - 6 x = x + 9$ $6x^2 - 6x = x + 9$ ?

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Answer 1 · 2015-05-28T12:23:03

Rewrite $6 x^{2} - 6 x = x + 9$ $6x^2-6x = x+9$
into the standard $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ form:

$6 x^{2} - 7 x - 9 = 0$ $6x^2-7x-9=0$

The quadratic formula for roots is
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

with the given example
$x = \frac{7 \pm \sqrt{49 + 216}}{12}$ $x= (7+-sqrt(49+216))/12$

$= \frac{7 \pm \sqrt{265}}{12}$ $= (7+-sqrt(265))/12$

How do you solve using the quadratic formula 6x2−6x=x+96x^2 - 6x = x + 9?