How do you solve #-2y^2+6y=-2#?

Question

1999 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-06T08:32:53

#color(blue)(y=(3+sqrt13)/(2) , y=(3-sqrt13)/(2#

#2y^2 -6y-2 =0#

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

#color(blue)(a=2, b=-6, c=-2#

The Discriminant is given by:
#Delta=b^2-4*a*c#

# = (-6)^2-(4*(2)*-2)#

# = 36 + 16=52#

The solution is found using the formula:

#y=(-b+-sqrtDelta)/(2*a)#

#y= (-(-6)+-sqrt(52))/(2*2) = (6+-sqrt(52))/4= (6+-2sqrt13)/4 #

#y=(6+2sqrt13)/(4) , x=(6-2sqrt13)/(4#

#color(blue)(y=(3+sqrt13)/(2) , y=(3-sqrt13)/(2#