Is this equation solvable for y?

[examorph]

I have been trying to solve the following equation for the variable y:
https://imageshack.com/i/0r9khg
Where K and P are constants, I tried an online solver which gave me the following result:
https://imageshack.com/i/nq4qm6g

I attempted to solve for this by adding the -1/2(n^2 t^2 y) to both sides and then dividing the -1/2(n^2 t^2) from both sides which left me with a big mess on the right, were the steps followed so far correct? Where would I go from here to solve for y?

Also, it has been suggested to me that this may require the use of the quadratic formula, is this correct?

Thanks.

 

[jedishrfu]

Just move the -1/2(n^2 t^2 y) to the left side and factor out the y and from there it should be obvious...

 

[examorph]

Thanks for the reply. I have tried this:
https://imageshack.com/i/mu2vb0g
When I divide both sides by the coefficient I get a large formula on the right hand side which does not match the final result shown by the solver, I had also been told that the quadratic formula is required to solve this, could you please show me what I must do next?

 

[Ray Vickson]

I have been trying to solve the following equation for the variable y:
https://imageshack.com/i/0r9khg
Where K and P are constants, I tried an online solver which gave me the following result:
https://imageshack.com/i/nq4qm6g

I attempted to solve for this by adding the -1/2(n^2 t^2 y) to both sides and then dividing the -1/2(n^2 t^2) from both sides which left me with a big mess on the right, were the steps followed so far correct? Where would I go from here to solve for y?

Also, it has been suggested to me that this may require the use of the quadratic formula, is this correct?

Thanks.

You have a simple equation of the form $y = A - By$, where
A = \frac{kt}{p}-\frac{k \sin(pt)}{p^2} \text{ and } B = \frac{n^2 t^2}{2}
Solving for y ought to be simple for you, without recourse to an on-line solver. Anyway, the solution given to you by the on-line solver is wrong. Are you sure you submitted the right equation?