How do you find the roots, real and imaginary, of $y = 7 x^{2} + 4 x + 6$ $y=7x^2+4x + 6$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = 7 x^{2} + 4 x + 6$ $y=7x^2+4x + 6$ using the quadratic formula?

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Answer 1 · 2016-05-06T13:47:44

$\frac{- 2 \pm i \sqrt{38}}{7}$ $(-2 +- isqrt38)/7$

Explanation:

Use the improved quadratic formula (Google, Yahoo Search)
y = 7x^2 + 4x + 6 = 0
D = d^2 = b^2 - 4ac = 16 - 168 = - 152
Since D < 0, there are no real roots. There are 2 complex roots.
$d^{2} = 152 i^{2}$ $d^2 = 152i^2$ --> $d = \pm 2 i \sqrt{38}$ $d = +- 2isqrt38$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{4}{14} \pm \frac{2 i \sqrt{38}}{14} = \frac{- 2 \pm i \sqrt{38}}{7}$ $x = -b/(2a) +- d/(2a) = -4/14 +- (2isqrt38)/14 = (-2 +- isqrt38)/7$

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How do you find the roots, real and imaginary, of $y = 7 x^{2} + 4 x + 6$ $y=7x^2+4x + 6$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=7x^2+4x + 6y=7x2+4x+6y=7x^2+4x + 6 using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of $y = 7 x^{2} + 4 x + 6$ $y=7x^2+4x + 6$ using the quadratic formula?