Solve square root of( x-2) =5

Solve square root of( x-2) =5

Solve square root of( 2x^2-7)=3 – x

Solve 3 square root of (y-7 ) +10(outside of square root) =2

Solve square root of (x+5) – square root of (x+7) = -10

if √(x-2)=5
then by squaring both sides you get:
x-2=25
Then add 2 to each side
x=27

if √(2x^2-7)=3 – x
then by squaring both sides again you get:
2x^2 - 7 = (3 - x)(3 - x)
expand out the brackets:
2x^2 - 7 = 9 - 6x + x^2
Make equal to zero:
x^2 + 6x - 16 = 0
Then just factories:
(x - 2)(x + 8)=0
Therefore x either equals 2 or -8

The same sort of processes should work for the others :)

(x-2) = 25
x = 27

2x^2 - 7 = 9 -6x +x^2
x^2 +6x -16 = 0
(x+8)(x-2) = 0
x = -8 or 2

Solve rest yourself