Question 134721
You have the correct steps. 


All that you need to do is solve for x in {{{x^2-2x+10}}}



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



Starting with the general quadratic


{{{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 {{{x^2-2*x+10=0}}} ( notice {{{a=1}}}, {{{b=-2}}}, and {{{c=10}}})





{{{x = (--2 +- sqrt( (-2)^2-4*1*10 ))/(2*1)}}} Plug in a=1, b=-2, and c=10




{{{x = (2 +- sqrt( (-2)^2-4*1*10 ))/(2*1)}}} Negate -2 to get 2




{{{x = (2 +- sqrt( 4-4*1*10 ))/(2*1)}}} Square -2 to get 4  (note: remember when you square -2, you must square the negative as well. This is because {{{(-2)^2=-2*-2=4}}}.)




{{{x = (2 +- sqrt( 4+-40 ))/(2*1)}}} Multiply {{{-4*10*1}}} to get {{{-40}}}




{{{x = (2 +- sqrt( -36 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (2 +- 6*i)/(2*1)}}} 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 = (2 +- 6*i)/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{x=1 + 3*i}}} or {{{x=1 - 3*i}}}




So the remaining zeros are 


{{{x=1 + 3*i}}} or {{{x=1 - 3*i}}}