How do you find the zeros of $y = 4/5x^2 - 7/2x + 2/3$ using the quadratic formula?

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How do you find the zeros of $y = 4/5x^2 - 7/2x + 2/3$ using the quadratic formula?

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Answer 1 · 2018-05-26T11:45:28

$x_1=35/16+sqrt(9195)/48$ or $x_2=35/16-sqrt(9105)/48$

Explanation:

Multiplying the whole equation by $30$
$24x^2-105x+20=0$
then we get
$x^2-105/24x+5/6=0$
$x_{1,2}=35/16pmsqrt((35/16)^2-5/6)$
$x_{1,2}=35/16pm\sqrt(3035/768)$
$x_{1,2}=35/16pm\sqrt(9105)/48$

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How do you find the zeros of $y = 4/5x^2 - 7/2x + 2/3$ using the quadratic formula?

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Science

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Humanities

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How do you find the zeros of y = 4/5x^2 - 7/2x + 2/3 y = 4/5x^2 - 7/2x + 2/3 using the quadratic formula?

1 Answer

Explanation:

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How do you find the zeros of $y = 4/5x^2 - 7/2x + 2/3$ using the quadratic formula?