Question 106313
Solve for x in terms of a:
1.) {{{3x^2+ax=a^2}}}  subtract {{{a^2}}} from both sides

{{{3x^2+ax-a^2=0}}}  quadratic in standard form.  
A=3
B=a
C=-a^2

We can solve using the quadratic formula:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-a +- sqrt( a^2-4*3*(-a^2 )))/(2*3) }}}

{{{x = (-a +- sqrt( a^2+12*a^2 ))/(6) }}}

{{{x = (-a +- sqrt( 13a^2 ))/(6) }}}
{{{x = (-a +- a*sqrt( 13))/(6) }}}
{{{x = a(-1 +- sqrt( 13 ))/(6) }}}

Check my work.


Hope this helps---ptaylor