How do you find the zeros, real and imaginary, of $y=5x^2+7x+12$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=5x^2+7x+12$ using the quadratic formula?

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Answer 1 · 2016-06-05T11:46:08

$x=-7/10+-sqrt(191)/10i$

Explanation:

let's solve the equation
$5x^2+7x+12=0$
using the quadratic formula, so
$x=(-7+-sqrt(49-240))/10$
under root we have a negative number, so the solutions are not real
$x=-7/10-sqrt(191)/10i$
and
$x=-7/10+sqrt(191)/10i$

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How do you find the zeros, real and imaginary, of $y=5x^2+7x+12$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of y=5x^2+7x+12y=5x^2+7x+12 using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=5x^2+7x+12$ using the quadratic formula?