How do you solve $3x^2 - 2x = 4$ using the quadratic formula?

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Answer 1 · 2016-02-27T04:10:16

The solutions are :
$color(blue)(x=(1+sqrt13)/3$

$color(blue)(x=(1-sqrt13)/3$

$3x^2-2x-4=0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=3, b=-2, c=-4$

The Discriminant is given by:
$Delta=b^2-4*a*c$

$= (-2)^2-(4*3*-4)$

$= 4 +48 = 52$

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$x = (-(-2)+-sqrt(52))/(2*3) = (2+-sqrt(52))/6$

Upon further simplification $sqrt52= sqrt(2*2*13)= 2sqrt13$

So, $x = (2+-2sqrt(13))/6 = (cancel2(1+-sqrt13))/cancel6$
$= (1+-sqrt13)/3$

The solutions are :
$color(blue)(x=(1+sqrt13)/3$
$color(blue)(x=(1-sqrt13)/3$

How do you solve 3x^2 - 2x = 43x^2 - 2x = 4 using the quadratic formula?