# How do you solve 4x^2 - 12x - 3 = 0 by completing the square?

Jun 12, 2017

$x = \frac{3}{2} \pm \sqrt{3}$

$x \approx - 0.23 \mathmr{and} x \approx + 3.23$

#### Explanation:

Given:$\text{ } 0 = 4 {x}^{2} - 12 x - 3$

For a complete explanation of the method see:
https://socratic.org/s/aFvmbGAs

It uses different numbers.

Set $y = 0 = 4 {x}^{2} - 12 x - 3$

$y = 0 = 4 {\left(x - \frac{12}{4 \times 2}\right)}^{2} + k - 3$

$y = 0 = 4 {\left(x - \frac{3}{2}\right)}^{2} + k - 3$
.........................................................................

Set $\text{ } 4 {\left(- \frac{3}{2}\right)}^{2} + k = 0$

$4 \times \frac{9}{4} + k = 0 \text{ "=>" } k = - 9$
.......................................................................

$y = 0 = 4 {\left(x - \frac{3}{2}\right)}^{2} - 12$

$+ \frac{12}{4} = {\left(x - \frac{3}{2}\right)}^{2}$

$x - \frac{3}{2} = \pm \sqrt{3}$

$x = \frac{3}{2} \pm \sqrt{3}$
$x = - 0.232050 . . \mathmr{and} x = + 3.232 - 50. \ldots$