Question 173854
Step 1: Put your equation into standard form:  {{{ax^2+bx+c=0}}}


{{{4x^2=2x-11}}} → {{{4x^2-2x+11=0}}}


Now you have {{{a=4}}}, {{{b=-2}}}, and {{{c=11}}}


So, substitute for a, b, and c into the quadratic formula:


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


{{{x = (-(-2) +- sqrt( (-2)^2-4(4)(11) ))/(2(4)) }}} 


Then do the arithmetic:


{{{x = (2 +- sqrt( 4-176 ))/8) }}}


{{{x = (2 +- sqrt(-172 ))/8) }}}


Oops! We have a negative number under the radical, so we have to invoke the imaginary number <i>i</i> defined as {{{i^2=-1}}}.


{{{x = (2 +- sqrt((-1)*172 ))/8) }}}


{{{x = (2 +- sqrt(i^2*172))/8) }}}


And you need to find the factorization of 172:  {{{2*2*43=172}}}, so:


{{{x = (2 +- sqrt(i^2*4*43))/8) }}}


Hence:  {{{x[1]=(2+2i*sqrt(43))/8=(1+i*sqrt(43))/4}}} and {{{x[2]=(2-2i*sqrt(43))/8=(1-i*sqrt(43))/4}}}