Whakaoti i te (y+3)^2=-8(x+2)

$$( y + 3 ) ^ { 2 } = - 8 ( x + 2 )$$

Whakaoti mō x

$$x=-\frac{\left(y+3\right)^{2}}{8}-2$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(y+3\right)^{2}.

$$y^{2}+6y+9=-8\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -8 ki te x+2.

$$y^{2}+6y+9=-8x-16$$

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

$$-8x-16=y^{2}+6y+9$$

Me tāpiri te 16 ki ngā taha e rua.

$$-8x=y^{2}+6y+9+16$$

Tāpirihia te 9 ki te 16, ka 25.

$$-8x=y^{2}+6y+25$$

Whakawehea ngā taha e rua ki te -8.

$$\frac{-8x}{-8}=\frac{y^{2}+6y+25}{-8}$$

Mā te whakawehe ki te -8 ka wetekia te whakareanga ki te -8.

$$x=\frac{y^{2}+6y+25}{-8}$$

Whakawehe y^{2}+6y+25 ki te -8.

$$x=-\frac{y^{2}}{8}-\frac{3y}{4}-\frac{25}{8}$$

Whakaoti mō y (complex solution)

$$y=-2\sqrt{-2x-4}-3$$

$$y=2\sqrt{-2x-4}-3$$

Whakaoti mō y

$$y=-2\sqrt{-2x-4}-3$$

$$y=2\sqrt{-2x-4}-3\text{, }x\leq -2$$