How do you solve #3y^2 - 7y + 4 = 0# using the quadratic formula?

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Answer 1 · 2018-07-18T12:07:10

The solutions are #S={1,4/3}#

The quadratic equation is

#3y^2-7y+4=0#

The discriminant is

#Delta=b^2-4ac=(-7)^2-4*3*4=49-48=1#

As #Delta>0#, there are #2# real roots

Therefore,

#y=(-b+-sqrt(Delta))/(2a)#

#=(-(-7)+-sqrt1)/(2*3)#

#=(7+-1)/6#

Therefore,

#y_1=8/6=4/3#

and

#y_2=6/6=1#

The solutions are #S={1,4/3}#

Answer 2 · 2018-07-18T12:23:58

#y=4/3, 1#

Given quadratic equation:

#3y^2-7y+4=0#

#y=\frac{-(-7)\pm\sqrt{(-7)^2-4(3)(4)}}{2(3)}#

#y=\frac{7\pm1}{6}#

#y=4/3, 1#