Question 173854
To use the quadratic formula, get it into the form,
{{{ax^2+bx+c=0}}}
.
.
{{{4x^2=2x-11}}}
{{{4x^2-2x+11=0}}}
Now for the quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-(-2) +- sqrt( (-2)^2-4*4*11 ))/(2*4) }}} 
{{{x = (2 +- sqrt( 4-176))/(8) }}} 
{{{x = (2 +- sqrt( -172))/(8) }}} 
{{{x = (2 +- sqrt( 172)i)/(8) }}}
or approximately,
{{{x = (2 +- 13.114i)/(8) }}}
{{{x = 0.25 +- 1.639i }}}