How do you solve $2 x^{2} + 32 x + 12 = 0$ $2x^2 + 32x + 12 = 0$ using completing the square?

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How do you solve $2 x^{2} + 32 x + 12 = 0$ $2x^2 + 32x + 12 = 0$ using completing the square?

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Answer 1 · 2015-06-24T23:04:18

I found:
$x_{1} = - 8 + \sqrt{58}$ $x_1=-8+sqrt(58)$
$x_{2} = - 8 - \sqrt{58}$ $x_2=-8-sqrt(58)$

Explanation:

Write it as (dividing by $2$ $2$ ):
$\frac{x^{2}}{2} + \frac{32}{2} x + \frac{12}{2} = 0$ $x^2/2+32/2x+12/2=0$
$x^{2} + 16 x + 6 = 0$ $x^2+16x+6=0$
$x^{2} + 16 x = - 6$ $x^2+16x=-6$
Add and subtract $64$ $64$ :
$x^{2} + 16 x + 64 - 64 = - 6$ $x^2+16x+64-64=-6$
$x^{2} + 16 x + 64 = - 6 + 64$ $x^2+16x+64=-6+64$
${(x + 8)}^{2} = 58$ $(x+8)^2=58$
$x + 8 = \pm \sqrt{58}$ $x+8=+-sqrt(58)$
$x = - 8 \pm \sqrt{58}$ $x=-8+-sqrt(58)$
so:
$x_{1} = - 8 + \sqrt{58}$ $x_1=-8+sqrt(58)$
$x_{2} = - 8 - \sqrt{58}$ $x_2=-8-sqrt(58)$

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How do you solve $2 x^{2} + 32 x + 12 = 0$ $2x^2 + 32x + 12 = 0$ using completing the square?

1 Answer

Explanation:

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