Question 754279
<pre>

              3x = &#8730;<span style="text-decoration: overline">20 - 3x</span>

Square both sides of the equation:

           (3x)² = (&#8730;<span style="text-decoration: overline">20 - 3x</span>)²
 
            3²x² = 20 - 3x

             9x² = 20 - 3x

   9x² + 3x - 20 = 0

(3x - 4)(3x + 5) = 0

Use the zero-factor principle:

3x - 4 = 0;  3x + 5 = 0
    3x = 4;      3x = -5
     x = {{{4/3}}};      x = {{{(-5)/3}}}      
                  x = {{{-5/3}}}

We must check radical equations to see if answers 
are solutions or extraneous answers:

Checking x = {{{4/3}}} in original equation:

{{{3(4/3)=sqrt(20-3(4/3))}}}
{{{cross(3)(4/cross(3))=sqrt(20-cross(3)(4/cross(3)))}}}
{{{4=sqrt(20-4)}}}
{{{4=sqrt(16)}}}
{{{4=4}}}

So {{{4/3}}} is a solution.

Checking x = {{{-5/3}}} in original equation:

{{{3(-5/3)=sqrt(20-3(-5/3))}}}
{{{cross(3)(-5/cross(3))=sqrt(20-cross(3)(-5/cross(3)))}}}
{{{-5=sqrt(20-(-5))}}}
{{{-5=sqrt(20+5))}}}
{{{-5=sqrt(25)}}}
{{{-5=5}}}

So {{{-5/3}}} is not a solution, but an extraneous answer.

Therefore the only solution is {{{4/3}}}.

Edwin</pre>