How do you solve $12 y^{2} - 5 y = 2$ $12y^2-5y=2$ ?

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Answer 1 · 2016-05-20T11:46:11

The solution is:
$y = \frac{2}{3}, y = - \frac{1}{4}$ $color(blue)(y=2/3, y=-1/4$

$12 y^{2} - 5 y = 2$ $12y^2 - 5y =2$

$12 y^{2} - 5 y - 2 = 0$ $12y^2 - 5y - 2 = 0$

The equation is of the form $a y^{2} + b y + c = 0$ $color(blue)(ay^2+by+c=0$ where:

$a = 12, b = - 5, c = - 2$ $a=12, b=-5, c=-2$

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (-5)^2-(4* 12 * (-2))$

$= 25 + 96 = 121$

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

$y = (-(-5)+-sqrt(121))/(2*12) = (5+-11)/24$

$y = (5 + 11 ) / 24 = 16/24 = 2/3$

$y = (5 - 11 ) / 24 = -6/24 = -1/4$

How do you solve 12y^2-5y=212y2−5y=212y^2-5y=2?