How do you solve #4y = 2( y - 5) ^ { 2} - 2#?
1 Answer
Dec 8, 2016
Expand everything.
#4y = 2(y^2 - 10y + 25) - 2#
#4y = 2y^2 - 20y + 50 - 2#
#0 = 2y^2 - 24y + 48#
#0 = 2(y^2 - 12y + 24)#
#y = (-(-12) +- sqrt(-12^2 - 4 xx 1 xx 24))/(2 xx 1)#
#y = (12 +- sqrt(48))/2#
#y = (12 +- 4sqrt(3))/2#
#y = 6 +- 2sqrt(3)#
Hopefully this helps!
