Solving $3y + 2y^2 = -e^{-x} - e^x + c$ for $y$

Question

So, I am failing to understand some potentially simple algebra here. I have a separable equation:$$\frac{dy}{dx} = \frac{e^{-x} - e^x}{3+4y},$$

and after the easy integration I get$$3y + 2y^2 = -e^{-x} - e^x + c .$$

Now, how do I solve for $y$? The book has a fairly long answer involving a square root... It could come down to, I did the separable/integration part incorrectly or I've lost my mind but I'm kind of shaking my head over my lack of algebra skills.

Answer 1

This is a quadratic equation for $y$, so you can feed it to the quadratic formula. The result is $$y=\frac {-3 \pm \sqrt{9-4\cdot 2(c-e^{-x}-e^x)}}{4\cdot 2}$$

Answer 2

The equation

$$2y^2+3y +e^{-x} + e^x - c=0$$

is a quadratic in terms of $y$, right?

So, use Bhaskara's formula (also called Quadratic Equation) to solve for $y$.