Question 128762
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^2-6*x+16=0}}} ( notice {{{a=1}}}, {{{b=-6}}}, and {{{c=16}}})





{{{x = (--6 +- sqrt( (-6)^2-4*1*16 ))/(2*1)}}} Plug in a=1, b=-6, and c=16




{{{x = (6 +- sqrt( (-6)^2-4*1*16 ))/(2*1)}}} Negate -6 to get 6




{{{x = (6 +- sqrt( 36-4*1*16 ))/(2*1)}}} Square -6 to get 36  (note: remember when you square -6, you must square the negative as well. This is because {{{(-6)^2=-6*-6=36}}}.)




{{{x = (6 +- sqrt( 36+-64 ))/(2*1)}}} Multiply {{{-4*16*1}}} to get {{{-64}}}




{{{x = (6 +- sqrt( -28 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (6 +- 2*i*sqrt(7))/(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 = (6 +- 2*i*sqrt(7))/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{x=3 + sqrt(7)i}}} or {{{x=3 - sqrt(7)i}}}


Notice if we graph the quadratic {{{y=x^2-6*x+16}}}, we get


{{{ graph( 500, 500, -12, 18, -8, 22, x^2-6*x+16) }}} graph of {{{y=x^2-6*x+16}}}


And we can see that there are no real roots