Question 344472
I can't decipher your question.
Use the caret to denote exponentiation, x^2={{{x^2}}}
Use sqrt(x) to denote the square root of x, sqrt(x)={{{sqrt(x)}}}
Is this what you mean??
{{{root(4,(x^2+6x))=2}}}
{{{x^2+6x=2^4}}}
{{{x^2+6x=16}}}
{{{x^2+6x-16=0}}}
{{{(x+8)(x-2)=0}}}
Two solutions:
{{{x+8=0}}}
{{{highlight(x=-8)}}}
.
.
{{{x-2=0}}}
{{{highlight(x=2)}}}