How do you find the zeros, real and imaginary, of $y=2x^2-5x+13$ using the quadratic formula?

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Answer 1 · 2015-11-22T00:38:08

$x=5/4+-sqrt(79)/4 i$

$2x^2-5x+13$ is in the form $ax^2+bx+c$ with $a=2$ , $b=-5$ and $c=13$ .

This has zeros given by the quadratic formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$=(-(-5)+-sqrt((-5)^2-(4xx2xx13)))/(2*2)$

$=(5+-sqrt(25-104))/4$

$=5/4+-sqrt(-79)/4$

$=5/4+-sqrt(79)/4 i$

How do you find the zeros, real and imaginary, of y=2x^2-5x+13 y=2x^2-5x+13 using the quadratic formula?