Question 275259
The zeros are the x values which make the equation {{{2x^2-x+4=0}}} true.





{{{2x^2-x+4=0}}} Start with the given equation.



Notice that the quadratic {{{2x^2-x+4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=-1}}}, and {{{C=4}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-1) +- sqrt( (-1)^2-4(2)(4) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=-1}}}, and {{{C=4}}}



{{{x = (1 +- sqrt( (-1)^2-4(2)(4) ))/(2(2))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(2)(4) ))/(2(2))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-32 ))/(2(2))}}} Multiply {{{4(2)(4)}}} to get {{{32}}}



{{{x = (1 +- sqrt( -31 ))/(2(2))}}} Subtract {{{32}}} from {{{1}}} to get {{{-31}}}



{{{x = (1 +- sqrt( -31 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (1 +- i*sqrt(31))/(4)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



Recall that 'i' is the imaginary number and {{{i=sqrt(-1)}}}



{{{x = (1+i*sqrt(31))/(4)}}} or {{{x = (1-i*sqrt(31))/(4)}}} Break up the expression.  



So the zeros are {{{x = (1+i*sqrt(31))/(4)}}} or {{{x = (1-i*sqrt(31))/(4)}}} 



Note: if you've never seen imaginary numbers before, then the answer is simply "no solutions"