How do you solve #2y^2 + 10y = -11 #?

1 Answer
Jul 7, 2016

Answer:

#y =-1.634 " or " y = -3.366#

Explanation:

As it is a quadratic, first make it = 0.
#2y^2 +10y +11 = 0 , " "ax^2 + bx + c =0#

There are 3 options:
factorise - #" "#(this quadratic trinomial does not factorise)
completing the square (but a = 2, )
quadratic formula feels best for this one

#y = (-b +-sqrt(b^2-4ac))/(2a) " with " a = 2, b= 10, c = 11#

#y = (-(10) +-sqrt(10^2-4(2)(11)))/(2(2))#

#y = (-10 +-sqrt(100-88))/4#

#y = (-10 +-sqrt12)/4#

We now solve it twice - once with + root and once with - root.
#y = (-10 +sqrt12)/4 " or "y = (-10 -sqrt12)/4#

#y =-1.634 " or " y = -3.366#