How to solve an equation with powers in it: $16y+4y^2=-11$

Question

Hello I'm confused as to how I simplify this equation or if it's even possible. $$16y+4y^2=-11$$ I was trying to get the logarithm of everything so I could get the exponent as a factor but it didn't seem to work.

Answer 1

In the general case of a second-order equation, for $y \in \mathbb{R}$:

$$ay^2+by+c=0$$ $$\Delta=b^2-4ac; \Delta \geq 0$$ $$y_{±}=\dfrac{-b±\sqrt{\Delta}}{2a}$$

Answer 2

It is equivalent to $$ 4y^2 + 16y + 11 = 0 $$ Now this is a quadratic equation.