Question 44135
Rearrange so you have zero on the right hand side:
{{{4x^2+5 = 3x}}}
{{{4x^2-3x+5 = 0}}}
Now use the quadratic solver {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
where
a=4
b=-3
c=5
This gives:
{{{x = (3 +- sqrt( 3^2-4*4*5 ))/(2*4) }}}
{{{x = (3 +- sqrt( 9-80 ))/8 }}}
{{{x = (3 +- sqrt( -71 ))/8 }}}
{{{x = (3 +- sqrt( 71 )i)/8 }}}
where "i" denotes the imaginary number {{{i=sqrt(-1)}}}

I hope this helps 
P.S. I am trying to start up my own homework help website. I would be extremely grateful if you would e-mail me some feedback on the help you received to adam.chapman@student.manchester.ac.uk