n+2=sqrt{a-n} eching

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

a uchun yechish

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

$$n+2\geq 0$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

Tenglamaning ikkala tarafidan -n ni ayirish.

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

O‘zidan -n ayirilsa 0 qoladi.

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

\left(n+2\right)^{2} dan -n ni ayirish.

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

n uchun yechish

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

$$a\geq -2$$