How do you solve the quadratic using the quadratic formula given $- c^{2} - 6 c + 8 = 0$ $-c^2-6c+8=0$ over the set of complex numbers?

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How do you solve the quadratic using the quadratic formula given $- c^{2} - 6 c + 8 = 0$ $-c^2-6c+8=0$ over the set of complex numbers?

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Answer 1 · 2018-01-02T12:45:14

$c = \frac{- 6 \pm 2 \sqrt{17}}{2}$ $c=(-6+-2sqrt(17))/2$

Explanation:

The quadratic formula states that if you have an equation in the form:
$n x^{2} + p x + q = 0$ $nx^2+px+q=0$

The solution(s) will be:
$x = \frac{- p \pm \sqrt{p^{2} - 4 n q}}{2 n}$ $x=(-p+-sqrt(p^2-4nq))/(2n)$

In our case $n = - 1, p = - 6$ $n=-1,p=-6$ and $q = 8$ $q=8$ , and applying the quadratic formula gives us:
$c = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 \cdot - 1 \cdot 8}}{2 \cdot - 1}$ $c=(-(-6)+-sqrt((-6)^2-4*-1*8))/(2*-1)$

$c = \frac{6 \pm \sqrt{36 - (- 32)}}{- 2}$ $c=(6+-sqrt(36-(-32)))/-2$

$c = - \frac{6 \pm \sqrt{36 + 32}}{2}$ $c=-(6+-sqrt(36+32))/2$

$c = \frac{- 6 \pm \sqrt{68}}{2}$ $c=(-6+-sqrt(68))/2$

$c = \frac{- 6 \pm \sqrt{2 \cdot 2 \cdot 17}}{2}$ $c=(-6+-sqrt(2*2*17))/2$

$c = \frac{- 6 \pm 2 \sqrt{17}}{2}$ $c=(-6+-2sqrt(17))/2$

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How do you solve the quadratic using the quadratic formula given $- c^{2} - 6 c + 8 = 0$ $-c^2-6c+8=0$ over the set of complex numbers?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve the quadratic using the quadratic formula given −c2−6c+8=0-c^2-6c+8=0 over the set of complex numbers?

1 Answer

Explanation:

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Impact of this question

How do you solve the quadratic using the quadratic formula given $- c^{2} - 6 c + 8 = 0$ $-c^2-6c+8=0$ over the set of complex numbers?