Question 82632
{{{2^x^2 - 2x = 8}}} and 
{{{3^x^2 + 4x = 1/81}}}
For the 1st one,
subtract 8 from both sides
{{{2x^2 - 2x -8 = 0}}}
divide both sides by 2
{{{x^2 - x - 4 = 0}}}
The easiest way is to complete the square
Take half the coefficient of x, square it, then add that
to both sides
To set it up, I'll add 4 to both sides
{{{x^2 - x = 4}}}
{{{((-1) / 2)^2 = 1/4}}} add this to both sides
{{{x^2 - x + 1/4 = 4 + 1/4}}}
Notice the left side is a perfect square
{{{(x - 1/2)^2 = 17/4}}}
Take the sqrt of both sides
{{{x - 1/2 = (17)^(1/2) / 2}}}
{{{x = ((17)^(1/2)+ 1) / 2}}}
You can check this by plugging it back into the equation for x
------------------------------
{{{3x^2 + 4x = 1/81}}}
divide both sides by 3
{{{x^2 + (4/3)x = 1/3*81}}}
Complete the square again
{{{x^2 + (4/3)x + 4/9 = 1/3*81 + 4/9}}}
{{{x^2 + (4/3)x + 4/9 = (1 + 27*4) / 3*81}}}
{{{x + (2/3))^2 = 109 / 243}}}
{{{x + (2/3) = (109/243)^(1/2)}}}
{{{x = (109/243)^(1/2) - (2/3)}}}
I could have made a mistake, but that's the method