## Answer :

so you square both sides and get rid of the square root

x+7=(x-5)^2

(x-5)^2=x^2-10x+25

x+7=x^2-10x+25

subtract 7 from both sides

x=x^2-10x+18

subtract x from both sides

0=x^2-11x+18

so if xy=0 we can assume that x or/and y =0

factor out x^2-11+18

(find what two numbers multiply to get 18 and add to get -11)

-2 times -9=18

-2+(-9)=-11

(x-2)(x-9)=0

set them to zero

x-2=0

x=2

x-9=0

x=9

there are two answers -2 and -9

The radicand must be greater than or equal to 0.

[tex]x+7 \geq 0 \\ x \geq -7[/tex]

The value of the square root must be greater than or equal to 0.

[tex]x-5 \geq 0 \\ x \geq 0[/tex]

Therefore x≥5.

[tex]\sqrt{x+7}=x-5 \\ (\sqrt{x+7})^2=(x-5)^2 \\ x+7=x^2-10x+25 \\ 0=x^2-11x+18 \\ 0=x^2-9x-2x+18 \\ 0=x(x-9)-2(x-9) \\ 0=(x-2)(x-9) \\ x-2=0 \ \lor \ x-9=0 \\ x=2 \ \lor \ x=9[/tex]

2<5 so it's not a correct solution.

9≥5 so it's a correct solution.

The answer:

**x=9**