How do you solve $x^{2} + 3 x - 5 = 0$ $x^2 + 3x - 5 = 0$ by completing the square?

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Answer 1 · 2017-04-06T16:06:48

$x = - \frac{3}{2} \pm \frac{\sqrt{29}}{4}$ $x=-3/2+-sqrt29/4$

Add $5$ $5$ on both sides:

$x^{2} + 3 x = 5$ $x^2+3x=5$

Now add ${(\frac{b}{2} a)}^{2}$ $(b/2a)^2$ on both sides where $a = 1$ $a=1$ and $b = 3$ $b=3$

$x^{2} + 3 x + \frac{9}{4} = 5 + \frac{9}{4}$ $x^2+3x+9/4=5+9/4$

Rewrite:

${(x + \frac{3}{2})}^{2} = \frac{29}{4}$ $(x+3/2)^2=29/4$

Square root both sides. Don't forget that it's $\pm$ $+-$ the square root:

$x + \frac{3}{2} = \pm \sqrt{\frac{29}{4}}$ $x+3/2=+-sqrt(29/4)$

Subtract $\frac{3}{2}$ $3/2$ on both sides:

$x = - \frac{3}{2} \pm \sqrt{\frac{29}{4}}$ $x=-3/2+-sqrt(29/4)$

How do you solve x^2 + 3x - 5 = 0x2+3x−5=0x^2 + 3x - 5 = 0 by completing the square?