Black-Scholes Formula Satisfies Black-Scholes Equation

My professor told me that each term of the Black-Scholes formula satisfy the Black-Scholes PDE. I have been trying to work this out for the last week and have been unsuccessful. I correctly showed the first term satisfied it, but I have had difficulty with the second term. My work is below:

We need to show that:
$$Q=-ke^{-r(T-t)}N\bigg(\frac{log\big(\frac{S}{k}\big)+(r-\delta+\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}-\sigma\sqrt{T-t}\bigg)$$

(where $N(w)$ is a standard normal distribution evaluated at $w$) satisfies the following:
$$(r-\delta)Q_sS+Q_t+\frac{1}{2}Q_{ss}\sigma^2S^2-rQ=0$$

Where $Q_s,\;Q_{ss}\;and\,Q_t$ are the derivative of Q w.r.t S, the second derivative of Q w.r.t S, and the derivative of Q w.r.t t respectively.

Let $d_2=\frac{log\big(\frac{S}{k}\big)+(r-\delta+\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}-\sigma\sqrt{T-t}$

$$Q_s=\frac{-ke^{-r(T-t)}N'(d_2)}{s\sigma\sqrt{T-t}}$$
$$Q_{ss}=\frac{-ke^{-r(T-t)}N”(d_2)\big(\frac{1}{s\sigma\sqrt{T-t}}\big)-\sigma\sqrt{T-t}}{S^2\sigma^2(T-t)}$$
$$Q_t=-kre^{-r(T-t)}N(d_2)+N'(d2)\bigg(\frac{log\big(\frac{S}{k}\big)}{2\sigma(T-t)^{\frac{3}{2}}}-\frac{r-\delta+\frac{1}{2}\sigma^2+\sigma}{2\sqrt{T-t}}\bigg)$$

When I plug these derivatives into the Black-Scholes PDE, I cannot get the equality to hold. Can anyone please help?

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