Question 242782
how do you solve square root and the 3 outside of it but in the little root thing and then inside the root 4+6t-the 3 in the outside of the root (but on it) square root of 1-8t=0
<pre>
I think this is what you are describing:

{{{root(3,4+6t)-root(3,1-8t)=0}}}

Add the opposite of second term to both sides:

{{{root(3,4+6t)=root(3,1-8t)}}}

Raise both sides to the 3 power:

{{{(root(3,4+6t))^3=(root(3,1-8t))^3}}}

When you raise a root to the same power as its index,
you take away the root and the exponent and just
leave what's underneath:

{{{4+6t=1-8t}}}

{{{14t=-3}}}

{{{t=(-3)/14}}}

{{{t=-3/14}}}

Edwin</pre>