Question 193074
I'm assuming that you want to solve for "x" right?




{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-3) +- sqrt( (-3)^2-4(2)(2a) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-3}}}, and {{{c=2a}}}. Note: there are two different "a" terms here...



{{{x = (3 +- sqrt( (-3)^2-4(2)(2a) ))/(2(2))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(2)(2a) ))/(2(2))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9-16a ))/(2(2))}}} Multiply {{{4(2)(2a)}}} to get {{{16a}}}



{{{x = (3 +- sqrt( 9-16a ))/(4)}}} Multiply 2 and 2 to get 4



{{{x = (3 + sqrt( 9-16a ))/4}}} or {{{x = (3 - sqrt( 9-16a ))/4}}} Break up the "plus/minus" to form two equations.



So the solutions are {{{x = (3 + sqrt( 9-16a ))/4}}} or {{{x = (3 - sqrt( 9-16a ))/4}}}



If we knew the value of "a", then we could continue simplifying.