# Finding the radian of a second-order eqn

I am really confused on what to do with the value d This is the solution I have now but not sure if it is correct or not.

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But 112.64 is in degrees, not radians... – kennytm Oct 3 '12 at 9:52
@KennyTM so the radians I got is 1.97 – JackyBoi Oct 3 '12 at 12:52
First, you're being very careless in your use of the equals sign. It's not true that $3y_{n+2}+4y_{n+1}+9y_n=3m^2+4m+9$, and it's not true that $3m^2+4m+9=14$. Second, $\theta=\cos^{-1}(-4/(2\sqrt{27}))$ comes from out of the blue --- presumably, you are using some formula someone gave you, but it's not in evidence, so who knows whether it's right? But the $\sqrt{27}$ looks very suspicious. I suspect it has something to do with the $b^2-4ac=-92$, but $92=4\times23\ne4\times27$. Don't worry about $d$ --- it affects the particular, not the complementary, solution. – Gerry Myerson Oct 4 '12 at 7:34
@GerryMyerson yes submitted my answer and got it correct but tks for the help, about me being careless sorry about that thats just how I work out (my brain thinks like that) but totally agree with you that I cant expect people to read my mind, will present it in a nice manner. – JackyBoi Oct 5 '12 at 3:24
So, does the correct answer really have a $\sqrt{27}$ in it, not a $\sqrt{23}$? – Gerry Myerson Oct 5 '12 at 4:51