How do you solve #2y² + 2y = 1# using the quadratic formula?

1 Answer
Jun 23, 2016

Answer:

#y=-1/2+-sqrt(3)/2#
(see below for use of quadratic formula)

Explanation:

#2y^2+2y=1# is equivalent to #color(red)(2)y^2+color(blue)(2)ycolor(green)(-1)#

Using the general quadratic:
#color(white)("XXX")color(red)(a)y^2+color(blue)(b)y+color(green)(c)=0#
and applying the general quadratic formula
#color(white)("XXX")y=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4(color(red)(a))(color(green)(c))))/(2(color(red)(a))#

We have
#color(white)("XXX")y=(-color(blue)(2)+-sqrt(color(blue)(2)^2-4(color(red)(2))(color(green)(-1))))/(2(color(red)(2))#

#color(white)("XXX")y=(-2+-sqrt(12))/4=-1/2+-sqrt(3)/2#