Question 88034
Let's use the quadratic formula to solve for x:

Starting with the general quadratic


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


the general form of the quadratic equation is:


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


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


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




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




{{{x = (2 +- sqrt( 4+-12 ))/(2*3)}}} Multiply {{{-4*1*3}}} to get {{{-12}}}




{{{x = (2 +- sqrt( -8 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (2 +- 2*i*sqrt(2))/(2*3)}}} Simplify the square root (note: since we cannot take the square root of a negative value, we must factor out {{{i=sqrt(-1)}}} to make the radicand positive)




{{{x = (2 +- 2*i*sqrt(2))/(6)}}} Multiply 2 and 3 to get 6




After simplifying, the quadratic has roots of


{{{x=0.333333333333333 + 0.471404520791032i}}} or {{{x=0.333333333333333 - 0.471404520791032i}}}


Notice if we graph the quadratic {{{y=3*x^2-2*x+1}}}, we get


{{{ graph( 500, 500, -14.6666666666667, 15.3333333333333, -14.3333333333333, 15.6666666666667, 3*x^2-2*x+1) }}} graph of {{{y=3*x^2-2*x+1}}}


we can see that the quadratic has no x-intercepts. This means there are no real solutions. So this verifies our answer.