How do you solve the equation $3 x^{2} - 5 x - 8 = 2 x - 3$ $3x^2-5x-8=2x-3$ by completing the square?

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How do you solve the equation $3 x^{2} - 5 x - 8 = 2 x - 3$ $3x^2-5x-8=2x-3$ by completing the square?

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Answer 1 · 2017-07-24T11:25:08

$x = \frac{1}{6} (7 \pm \sqrt{109})$ $x =1/6(7+-sqrt109)$

Explanation:

$3 x^{2} - 5 x - 8 = 2 x - 3 or 3 x^{2} - 5 x - 8 - 2 x + 3 = 0$ $3x^2 - 5x -8 =2x-3 or 3x^2 - 5x -8 -2x+3 =0$ or

$3 x^{2} - 7 x - 5 = 0 or 3 (x^{2} - \frac{7}{3} x) - 5$ $3x^2 - 7x -5 =0 or 3 (x^2-7/3x) -5$ or

$3 (x^{2} - \frac{7}{3} x + {(\frac{7}{6})}^{2}) - 3 \cdot \frac{49}{36} - 5 = 0$ $3(x^2-7/3x+(7/6)^2) - 3*49/36 -5 =0$

$3 {(x - \frac{7}{6})}^{2} - \frac{49}{12} - 5 or 3 {(x - \frac{7}{6})}^{2} = \frac{109}{12}$ $3 (x-7/6)^2 - 49/12-5 or 3 (x-7/6)^2 = 109/12$ or

${(x - \frac{7}{6})}^{2} = \frac{109}{3 \cdot 12} or (x - \frac{7}{6}) = \pm \sqrt{\frac{109}{36}}$ $(x-7/6)^2 = 109/(3*12) or (x-7/6) = +- sqrt (109/36)$ or

$x = \frac{7}{6} \pm \frac{\sqrt{109}}{6} = \frac{1}{6} (7 \pm \sqrt{109})$ $x = 7/6+- sqrt (109)/6 =1/6(7+-sqrt109)$ [Ans]

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How do you solve the equation $3 x^{2} - 5 x - 8 = 2 x - 3$ $3x^2-5x-8=2x-3$ by completing the square?

1 Answer

Explanation:

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How do you solve the equation $3 x^{2} - 5 x - 8 = 2 x - 3$ $3x^2-5x-8=2x-3$ by completing the square?