Question

(radical 2x-8)+10=6

Answer 1

\(
\sqrt{2x-8} \ or\ (2x-8)^2?
\)

Answer 2

\[\sqrt{(2x - 8)} + 10 = 6\]
\[\sqrt{(2x - 8)} = 6-10 = -4\]

Now square both sides
\[(2x - 8) = 16\]
\[2x = 16 + 8 = 24\]
\[x = 24/2 = 12\]

now substitute the value in the original equation to see if it satisfies the condition.