Whakaoti i te 64=n+n(n-1)/2d

$$64 = n + \frac { n ( n - 1 ) } { 2 } d$$

Whakaoti mō d

$$d=-\frac{2\left(n-64\right)}{n\left(n-1\right)}$$

$$n\neq 1\text{ and }n\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te 2.

$$128=2n+n\left(n-1\right)d$$

Whakamahia te āhuatanga tohatoha hei whakarea te n ki te n-1.

$$128=2n+\left(n^{2}-n\right)d$$

Whakamahia te āhuatanga tohatoha hei whakarea te n^{2}-n ki te d.

$$128=2n+n^{2}d-nd$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$2n+n^{2}d-nd=128$$

Tangohia te 2n mai i ngā taha e rua.

$$n^{2}d-nd=128-2n$$

Pahekotia ngā kīanga tau katoa e whai ana i te d.

$$\left(n^{2}-n\right)d=128-2n$$

Whakawehea ngā taha e rua ki te n^{2}-n.

$$\frac{\left(n^{2}-n\right)d}{n^{2}-n}=\frac{128-2n}{n^{2}-n}$$

Mā te whakawehe ki te n^{2}-n ka wetekia te whakareanga ki te n^{2}-n.

$$d=\frac{128-2n}{n^{2}-n}$$

Whakawehe 128-2n ki te n^{2}-n.

$$d=\frac{2\left(64-n\right)}{n\left(n-1\right)}$$

Whakaoti mō n (complex solution)

$$\left\{\begin{matrix}n=-\frac{\sqrt{d^{2}+508d+4}-d+2}{2d}\text{; }n=-\frac{-\sqrt{d^{2}+508d+4}-d+2}{2d}\text{, }&d\neq 0\\n=64\text{, }&d=0\end{matrix}\right.$$

Whakaoti mō n

$$\left\{\begin{matrix}n=-\frac{\sqrt{d^{2}+508d+4}-d+2}{2d}\text{; }n=-\frac{-\sqrt{d^{2}+508d+4}-d+2}{2d}\text{, }&d\leq -96\sqrt{7}-254\text{ or }\left(d\neq 0\text{ and }d\geq 96\sqrt{7}-254\right)\\n=64\text{, }&d=0\end{matrix}\right.$$