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

Jul 7, 2016

$y = - 1.634 \text{ or } y = - 3.366$

#### Explanation:

As it is a quadratic, first make it = 0.
$2 {y}^{2} + 10 y + 11 = 0 , \text{ } a {x}^{2} + b x + c = 0$

There are 3 options:
factorise - $\text{ }$(this quadratic trinomial does not factorise)
completing the square (but a = 2, )
quadratic formula feels best for this one

$y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \text{ with } a = 2 , b = 10 , c = 11$

$y = \frac{- \left(10\right) \pm \sqrt{{10}^{2} - 4 \left(2\right) \left(11\right)}}{2 \left(2\right)}$

$y = \frac{- 10 \pm \sqrt{100 - 88}}{4}$

$y = \frac{- 10 \pm \sqrt{12}}{4}$

We now solve it twice - once with + root and once with - root.
$y = \frac{- 10 + \sqrt{12}}{4} \text{ or } y = \frac{- 10 - \sqrt{12}}{4}$

$y = - 1.634 \text{ or } y = - 3.366$