How do you solve $3y^2+5y-4=0$ using the quadratic formula?

Question

3763 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-26T14:27:54

The solutions are $y = (-5+sqrt(73))/6 , y = (-5-sqrt(73))/6$

$3y^2 + 5y - 4 = 0$

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

$a=3, b=5, c= -4$

The Discriminant is given by:

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

$= (5)^2-(4* 3 * (-4))$

$= 25 + 48 = 73$

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

$y = ((-5)+-sqrt(73))/(2*3) = (-5+-sqrt(73))/6$

The solutions are:

How do you solve 3y^2+5y-4=03y^2+5y-4=0 using the quadratic formula?