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

1 Answer
Apr 26, 2016

Answer:

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

Explanation:

#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:

  • #color(green)(y = (-5+sqrt(73))/6#

  • #color(green)(y = (-5-sqrt(73))/6#