How do you solve #2y^2 – 6y – 8 = 0# using the quadratic formula?

Question

2701 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-17T16:49:34

#y=-1 "; " y=4#

Just use #y# instead of #x#

#color(brown)("Given: "x=2y^2-6y-8=0)#

#a=2"; "b=-6"; "c=-8#

#x=ay^2+by+c#

Where: #" "y=(-b+-sqrt(b^2-4ac))/(2a)#

#=> y=(+6+-sqrt(6^2-4(2)(-8)))/(2(2))#

#y= 6/4+- sqrt(100)/4 #

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)(y=6/4+10/4=16/4=4)#

#color(brown)(y=6/4-10/4 = -4/4=-1)#