How do you solve $- 2 x^{2} + 10 x = - 14$ $-2x^2+10x=-14$ by completing the square?

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Answer 1 · 2018-01-22T08:24:35

$- 2 x^{2} + 10 x = - 14$ $−2x^2+10x=−14$

$2 x^{2} - 10 x - 14 = 0$ $2x^2-10x-14=0$

dividing the entire equation by 2,
$x^{2} - 5 x - 7 = 0$ $x^2-5x-7=0$

now adding and subtract "half of the $x$ $x$ term squared",
which in this case is, ${(\frac{5}{2})}^{2}$ $(5/2)^2$

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

${(x - \frac{5}{2})}^{2} - 7 - (\frac{25}{4}) = 0$ $(x- 5/2)^2 -7-(25/4)=0$

${(x - \frac{5}{2})}^{2} - \frac{28}{4} - \frac{25}{4} = 0$ $(x- 5/2)^2 -28/4-25/4=0$

${(x - \frac{5}{2})}^{2} - \frac{53}{4} = 0$ $(x- 5/2)^2 -53/4=0$

${(x - \frac{5}{2})}^{2} = \frac{53}{4}$ $(x- 5/2)^2 =53/4$

$(x - \frac{5}{2}) = \pm \sqrt{\frac{53}{4}}$ $(x- 5/2) =+-sqrt(53/4)$

$(x - \frac{5}{2}) = \pm \frac{\sqrt{53}}{2}$ $(x- 5/2) =+-sqrt53/2$

$x = \pm \frac{\sqrt{53}}{2} + \frac{5}{2}$ $x=+-sqrt53/2+ 5/2$

$x = \frac{5 \pm \sqrt{53}}{2}$ $x=(5+-sqrt53)/2$

How do you solve -2x^2+10x=-14−2x2+10x=−14-2x^2+10x=-14 by completing the square?