Question 134123
To find the roots, set {{{y = 0}}}
You first have to multiply both sides by {{{2}}} in 
order to get rid of the {{{.5}}} in front of the {{{x^2}}} term
{{{.5x^2 - 3x + 19/2 = 0}}}
{{{x^2 - 6x + 19 = 0}}}
The rule is, take 1/2 of the coefficient of the x-term, then
square it, then add the result to both sides
{{{x^2 - 6x + (6/2)^2 + 19 = (6/2)^2}}}
Now subtract {{{19}}} from both sides
{{{x^2 - 6x + 9 = 9 - 19}}}
{{{(x - 3)^2 = -10}}}
Right away, I see that something squared is supposed 
to be negative, so the "something", which is {{{x - 3}}},
must be an imaginary quantity
Take the square root of both sides
{{{x - 3 = 0 +-sqrt(10)*i}}}
{{{x = 3 +-sqrt(10)i}}}
There are 2 answers, {{{x = 3 + sqrt(10)i}}} and {{{x = 3 - sqrt(10)i}}}
check answer:
{{{(3-sqrt(10)*i)^2 - 6*(3-sqrt(10)*i) + 19 = 0}}}
{{{9 - 6*sqrt(10)*i + (-10) - 18 + 6*sqrt(10)*i + 19 = 0}}}
{{{0 = 0}}}
OK