How do you solve $4 x^{2} + 3 = - 12 x$ $4x^2 + 3 = −12x$ ?

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How do you solve $4 x^{2} + 3 = - 12 x$ $4x^2 + 3 = −12x$ ?

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Answer 1 · 2016-03-11T14:27:51

$x = - \frac{3}{2} \pm \frac{\sqrt{6}}{2}$ $x = -3/2 +- sqrt6/2$

Explanation:

$y = 4 x^{2} + 12 x + 3 = 0$ $y = 4x^2 + 12x + 3 = 0$
Use the improved quadratic formula (Google, Yahoo Search)
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} .$ $x = -b/(2a) +- d/(2a).$
$D = d^{2} = b^{2} - 4 a c = 144 - 48 = 96$ $D = d^2 = b^2 - 4ac = 144 - 48 = 96$ --> $d = \pm 4 \sqrt{6}$ $d = +- 4sqrt6$
$x = - \frac{12}{8} \pm 4 \frac{\sqrt{6}}{8}$ $x = - 12/8 +- 4sqrt6/8$
$x = - \frac{3}{2} \pm \frac{\sqrt{6}}{2}$ $x = -3/2 +- sqrt6/2$

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How do you solve $4 x^{2} + 3 = - 12 x$ $4x^2 + 3 = −12x$ ?

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How do you solve 4x^2 + 3 = −12x4x2+3=−12x4x^2 + 3 = −12x?

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How do you solve $4 x^{2} + 3 = - 12 x$ $4x^2 + 3 = −12x$ ?