Question 144281
the quadratic equation is :

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

Given : {{{3y^2-4=0}}}
a = 3 , b = 0, c =-4
{{{x = (0 +- sqrt( 0-4*3*(-4) ))/(2*3) }}} 
{{{x = (0 +- sqrt( 48))/(6) }}} 
{{{x = (0 +- sqrt( 16*3) )/(6) }}} 
{{{x = (0 +- 4*sqrt(3) )/(6) }}} 
{{{x = (0 +- 2*sqrt(3) )/3 }}} 


4u2+3u=0
use the same process for the second part. Except this time a=4, b=3, c=0.
Shame too. it would be very easy to factor as
{{{u*(4u+3) =0}}}
Which yields u=0, and u=-3/4