How do you find the zeros, real and imaginary, of $y=3x^2-2x+3$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=3x^2-2x+3$ using the quadratic formula?

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Answer 1 · 2018-05-27T16:51:33

$x=-1/3+2/3sqrt(2)i$ or $x=-1/3-2/3sqrt(2)i$

Explanation:

Given
$3x^2-2x+3=0$
dividing by $3$
$x^2-2/3x+1=0$
Using the quadratic formula
$x_{1,2}=-1/3pmsqrt(1/9-9/9)$
this is
$x_{1,2}=-1/3pm\sqrt(8/9)i$
simplifying
$x_{1,2}=-1/3\pm2/3\sqrt(2)i$

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How do you find the zeros, real and imaginary, of $y=3x^2-2x+3$ using the quadratic formula?

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Humanities

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How do you find the zeros, real and imaginary, of y=3x^2-2x+3y=3x^2-2x+3 using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=3x^2-2x+3$ using the quadratic formula?