How do you find the zeros of $y = 7 x^{2} + x - 2$ $y = 7x^2 + x -2$ using the quadratic formula?

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How do you find the zeros of $y = 7 x^{2} + x - 2$ $y = 7x^2 + x -2$ using the quadratic formula?

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Answer 1 · 2015-12-20T18:28:32

The roots are $\frac{- 1 \pm \sqrt{57}}{14}$ $(-1 +- sqrt(57))/14$ .

Explanation:

You first calculate $Delta = b^2 - 4ac$ , which here equals to $1 - 4*7*(-2) = 57$ . It's positive so $y$ has two roots.

The quadratic formula says that the roots of a trinomial are $(-b +- sqrt(Delta))/(2a$ . So the two roots of $y$ are $(-1 - sqrt(57))/14$ and $(-1 + sqrt(57))/14$ .

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How do you find the zeros of $y = 7 x^{2} + x - 2$ $y = 7x^2 + x -2$ using the quadratic formula?

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How do you find the zeros of y = 7x^2 + x -2 y=7x2+x−2 y = 7x^2 + x -2 using the quadratic formula?

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How do you find the zeros of $y = 7 x^{2} + x - 2$ $y = 7x^2 + x -2$ using the quadratic formula?