What are the zeros of the quadratic function $f (x) = 6 x^{2} + 12 x - 7$ $f(x)=6x^2+12x-7$ ?

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What are the zeros of the quadratic function $f (x) = 6 x^{2} + 12 x - 7$ $f(x)=6x^2+12x-7$ ?

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Answer 1 · 2016-11-19T00:23:22

$x = - 1 \pm \frac{1}{6} \sqrt{78}$ $x = -1+-1/6 sqrt(78)$

Explanation:

$f (x) = 6 x^{2} + 12 x - 7$ $f(x) = 6x^2+12x-7$

is of the form $a x^{2} + b x + c$ $ax^2+bx+c$ with $a = 6$ $a=6$ , $b = 12$ $b=12$ and $c = - 7$ $c=-7$

We can find the zeros of $f (x)$ $f(x)$ using the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- 12 \pm \sqrt{12^{2} - 4 (6) (- 7)}}{2 (6)}$ $color(white)(x) = (-12+-sqrt(12^2-4(6)(-7)))/(2(6))$

$x = \frac{- 12 \pm \sqrt{144 + 168}}{12}$ $color(white)(x) = (-12+-sqrt(144+168))/12$

$x = \frac{- 12 \pm \sqrt{312}}{12}$ $color(white)(x) = (-12+-sqrt(312))/12$

$x = \frac{- 12 \pm \sqrt{4 \cdot 78}}{12}$ $color(white)(x) = (-12+-sqrt(4*78))/12$

$x = \frac{- 12 \pm 2 \sqrt{78}}{12}$ $color(white)(x) = (-12+-2sqrt(78))/12$

$x = - 1 \pm \frac{1}{6} \sqrt{78}$ $color(white)(x) = -1+-1/6 sqrt(78)$

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What are the zeros of the quadratic function $f (x) = 6 x^{2} + 12 x - 7$ $f(x)=6x^2+12x-7$ ?

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Explanation:

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What are the zeros of the quadratic function f(x)=6x2+12x−7f(x)=6x^2+12x-7?

1 Answer

Explanation:

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What are the zeros of the quadratic function $f (x) = 6 x^{2} + 12 x - 7$ $f(x)=6x^2+12x-7$ ?