How do you solve $3 x^{2} + 10 x - 2 = 0$ $3x^2 + 10x - 2 = 0$ by completing the square?

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How do you solve $3 x^{2} + 10 x - 2 = 0$ $3x^2 + 10x - 2 = 0$ by completing the square?

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Answer 1 · 2017-10-07T06:56:55

$x = \frac{\sqrt{31}}{9} - \frac{5}{3}$ $x=sqrt31/9-5/3$
$x = - \frac{\sqrt{31}}{9} - \frac{5}{3}$ $x=-sqrt31/9-5/3$

Explanation:

Given -

$3 x^{2} + 10 x - 2 = 0$ $3x^2+10x-2=0$

Take the constant term to right-hand side

$3 x^{2} + 10 x = 2$ $3x^2+10x=2$

Divide both sides by 3

$\frac{3 x^{2}}{3} + \frac{10 x}{3} = \frac{2}{3}$ $(3x^2)/3+(10x)/3=2/3$

$x^{2} + \frac{10}{3} x = \frac{2}{3}$ $x^2+10/3x=2/3$

Divide the coefficient of $x$ $x$ by2; square it and add it to both sides

$x^{2} + \frac{10}{3} x + \frac{100}{36} = \frac{2}{3} + \frac{100}{36} = \frac{24 + 100}{36} = \frac{124}{36} = \frac{31}{9}$ $x^2+10/3x+100/36=2/3+100/36=(24+100)/36=124/36=31/9$
${(x + \frac{10}{6})}^{2} = \frac{31}{9}$ $(x+10/6)^2=31/9$

$x + \frac{5}{3} = \pm \sqrt{\frac{31}{9}} = \pm \frac{\sqrt{31}}{3}$ $x+5/3=+-sqrt(31/9)= +-sqrt31/3$

$x = \frac{\sqrt{31}}{9} - \frac{5}{3}$ $x=sqrt31/9-5/3$
$x = - \frac{\sqrt{31}}{9} - \frac{5}{3}$ $x=-sqrt31/9-5/3$

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How do you solve $3 x^{2} + 10 x - 2 = 0$ $3x^2 + 10x - 2 = 0$ by completing the square?

1 Answer

Explanation:

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How do you solve $3 x^{2} + 10 x - 2 = 0$ $3x^2 + 10x - 2 = 0$ by completing the square?