```Question 96724
Let's use the quadratic formula to solve for x:

{{{ax^2+bx+c=0}}}

the general solution using the quadratic equation is:

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

So lets solve {{{5*x^2-x+3=0}}} ( notice {{{a=5}}}, {{{b=-1}}}, and {{{c=3}}})

{{{x = (--1 +- sqrt( (-1)^2-4*5*3 ))/(2*5)}}} Plug in a=5, b=-1, and c=3

{{{x = (1 +- sqrt( (-1)^2-4*5*3 ))/(2*5)}}} Negate -1 to get 1

{{{x = (1 +- sqrt( 1-4*5*3 ))/(2*5)}}} Square -1 to get 1  (note: remember when you square -1, you must square the negative as well. This is because {{{(-1)^2=-1*-1=1}}}.)

{{{x = (1 +- sqrt( 1+-60 ))/(2*5)}}} Multiply {{{-4*3*5}}} to get {{{-60}}}

{{{x = (1 +- sqrt( -59 ))/(2*5)}}} Combine like terms in the radicand (everything under the square root)

{{{x = (1 +- i*sqrt(59))/(2*5)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)

{{{x = (1 +- i*sqrt(59))/(10)}}} Multiply 2 and 5 to get 10

After simplifying, the quadratic has roots of

{{{x=1/10 + sqrt(59)/10*i}}} or {{{x=1/10 - sqrt(59)/10*i}}}

Notice if we graph the quadratic {{{y=5*x^2-x+3}}}, we get

{{{ graph( 500, 500, -14.9, 15.1, -12.05, 17.95, 5*x^2-x+3) }}} graph of {{{y=5*x^2-x+3}}}

And we can see that there are no real roots

To visually verify the answer, check out <a href=http://www.math.hmc.edu/funfacts/ffiles/10005.1.shtml>this page</a> to see a visual representation of imaginary roots
```