Question

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

Answer 1

`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`