Question 34173
Not too bad ... you just need to keep going ...

Problem 1:
{{{ 3x^2 - 80 = 0 }}}
{{{ 3x^2 = 80 }}}
{{{ 3x^2/3 = 80/3 }}}
{{{ x^2 = 80/3 }}}
now when you square root the fraction .. you will get a denominator with a square root in it ... That is a BIG NO-NO!!!
{{{ sqrt(x^2) = sqrt(80/3) }}}
{{{ x = sqrt(80)/sqrt(3) }}}
To rationalize the denominator we need to multiply the numerator and denominator by root3
{{{ x = (sqrt(80))(sqrt(3))/(sqrt(3))(sqrt(3)) }}}
{{{ x = sqrt(240)/3 }}}
{{{ x = 4(sqrt(15))/3 }}}
And yes ... it is +-
{{{ x = 4(sqrt(15))/3 }}}
Done.



Problem 2:
{{{ (2x-2)^2 = 9 }}}
yes ... root both sides
here you will have two roots... one positive ... one negative
{{{ 2x-2 = 3 }}} or {{{ 2x-2 = -3 }}}
ADD 2
{{{ 2x = 5 }}} or {{{ 2x = -1 }}}
Divide by 2
{{{ x = 5/2 }}} or {{{ x = -1/2 }}}
When solving for a given variable .... we work PEMDAS backwards ... doing the adding and subtracting first ... and then the multiplying and dividing