How do you solve $4 x^{2} - 8 = - 13 x$ $4x^2-8=-13x$ by completing the square?

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How do you solve $4 x^{2} - 8 = - 13 x$ $4x^2-8=-13x$ by completing the square?

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Answer 1 · 2015-05-02T22:16:47

$y = 4 x^{2} + 13 x - 8 = 0 .$ $y = 4x^2 + 13x - 8 = 0.$

$y = x^{2} + \frac{13 x}{4} - 2 = 0$ $y = x^2 + (13x)/4 - 2 = 0$
$= x^{2} + 13 \frac{x}{4} + \frac{169}{16} - \frac{169}{16} - 2 = 0$ $= x^2 + 13x/4 + 169/16 - 169/16 - 2 = 0$

${(x + \frac{13}{8})}^{2} = \frac{169}{64} + 2 = \frac{297}{64}$ $(x + 13/8)^2 = 169/64 + 2 = 297/64$

$x = - \frac{13}{8} + \frac{297}{64} =$ $x = -13/8 + 297/64 =$
$x = - \frac{13}{8} - \frac{297}{64} =$ $x = -13/8 - 297/64 =$

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How do you solve $4 x^{2} - 8 = - 13 x$ $4x^2-8=-13x$ by completing the square?

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How do you solve $4 x^{2} - 8 = - 13 x$ $4x^2-8=-13x$ by completing the square?