Whakaoti i te 80=x(x-1)d

$$80 = x ( x - 1 ) d$$

Whakaoti mō d

$$d=-\frac{80}{x\left(1-x\right)}$$

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

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$80=\left(x^{2}-x\right)d$$

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

$$80=x^{2}d-xd$$

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

$$x^{2}d-xd=80$$

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

$$\left(x^{2}-x\right)d=80$$

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

$$\frac{\left(x^{2}-x\right)d}{x^{2}-x}=\frac{80}{x^{2}-x}$$

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

$$d=\frac{80}{x^{2}-x}$$

Whakawehe 80 ki te x^{2}-x.

$$d=\frac{80}{x\left(x-1\right)}$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{d\left(d+320\right)}+d}{2d}$$

$$x=\frac{-\sqrt{d\left(d+320\right)}+d}{2d}\text{, }d\neq 0$$

Whakaoti mō x

$$x=\frac{\sqrt{d\left(d+320\right)}+d}{2d}$$

$$x=\frac{-\sqrt{d\left(d+320\right)}+d}{2d}\text{, }d>0\text{ or }d\leq -320$$