How do you solve $2 x^{2} + 7 x - 5 = 0$ $2x^2+7x-5=0$ by completing the square?

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How do you solve $2 x^{2} + 7 x - 5 = 0$ $2x^2+7x-5=0$ by completing the square?

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Answer 1 · 2017-07-04T03:05:36

$x = \frac{- 7 \pm \sqrt{89}}{4}$ $x = (-7 +- sqrt89)/4$

Explanation:

$2 x^{2} + 7 x = 5$ $2x^2 + 7x = 5$
$x^{2} + \frac{7 x}{2} = \frac{5}{2}$ $x^2 + (7x)/2 = 5/2$
$x^{2} + \frac{7 x}{2} + \frac{49}{16} = \frac{5}{2} + \frac{49}{16}$ $x^2 + (7x)/2 + 49/16 = 5/2 + 49/16$
${(x + \frac{7}{4})}^{2} = \frac{89}{16}$ $(x + 7/4)^2 = 89/16$
$x + \frac{7}{4} = \pm \frac{\sqrt{89}}{4}$ $x + 7/4 = +- sqrt89/4$
$x = - \frac{7}{4} \pm \frac{\sqrt{89}}{4} = \frac{- 7 \pm \sqrt{89}}{4}$ $x = - 7/4 +- sqrt89/4 = (- 7 +- sqrt89)/4$

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How do you solve $2 x^{2} + 7 x - 5 = 0$ $2x^2+7x-5=0$ by completing the square?

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