How do you solve $2 x^{2} + 3 x - 17 = 0$ $2x^2+3x-17=0$ by completing the square?

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Answer 1 · 2016-06-18T15:15:21

$x = \frac{- 3 \pm \sqrt{145}}{4}$ $x=(-3+-sqrt145)/4$

step by step

$2 x^{2} + 3 x - 17 = 0$ $2x^2+3x-17=0$

$2 x^{2} + 3 x = 17$ $2x^2+3x=17$

$x^{2} + \frac{3}{2} x = \frac{17}{2}$ $x^2+3/2x=17/2$

$x^{2} + \frac{3}{2} x + \frac{9}{16} = \frac{17}{2} + \frac{9}{16}$ $x^2+3/2x+9/16=17/2+9/16$

${(x + \frac{3}{4})}^{2} = \frac{145}{16}$ $(x+3/4)^2=145/16$

$x + \frac{3}{4} = \pm \frac{\sqrt{145}}{4}$ $x+3/4=+-sqrt145/4$

$x = \frac{- 3 \pm \sqrt{145}}{4}$ $x=(-3+-sqrt145)/4$

How do you solve 2x2+3x−17=02x^2+3x-17=0 by completing the square?