# How do you solve using the completing the square method x^2+10x-4=0?

Aug 1, 2016

$x = \pm \sqrt{29} - 5$

#### Explanation:

${x}^{2} + 10 x - 4 = {\left(x + a\right)}^{2} + b$

${\left(x + a\right)}^{2} + b = {x}^{2} + 2 a x + {a}^{2} + b$

You will find $2 a x = 10 x$ hence $a = 5$
${a}^{2} + b = - 4$
${5}^{2} + b = - 4$
$b = - 29$

Now you have: ${\left(x + 5\right)}^{2} - 29 = 0$
${\left(x + 5\right)}^{2} = 29$
$\left(x + 5\right) = \pm \sqrt{29}$
$\therefore x = \pm \sqrt{29} - 5$

Aug 1, 2016

$x = - 5 \pm \sqrt{29}$
${x}^{2} + 10 x - 4 = 0 \mathmr{and} {\left(x + 5\right)}^{2} - 25 - 4 = 0 \mathmr{and} {\left(x + 5\right)}^{2} = 29 \mathmr{and} \left(x + 5\right) = \pm \sqrt{29} \mathmr{and} x = - 5 \pm \sqrt{29}$[Ans]