Whakaoti i te n+2=sqrt{a-n}

$$n + 2 = \sqrt { a - n }$$

Whakaoti mō a

$$a=\left(n+1\right)\left(n+4\right)$$

$$n+2\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

$$\sqrt{a-n}=n+2$$

Pūruatia ngā taha e rua o te whārite.

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

Me tango -n mai i ngā taha e rua o te whārite.

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

Mā te tango i te -n i a ia ake anō ka toe ko te 0.

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

Tango -n mai i \left(n+2\right)^{2}.

$$a=\left(n+2\right)^{2}+n$$

Whakaoti mō n

$$n=\frac{\sqrt{4a+9}-5}{2}$$

$$a\geq -2$$