# How do you solve 2x^2 – 28x + 84 = 0 using completing the square?

Jul 3, 2015

Logical first step is dividing everything by $2$

#### Explanation:

$\to {x}^{2} - 14 x + 42 = 0$

Take half the $x$-coefficient and square that:
${x}^{2} - 14 x + {\left(- 7\right)}^{2} = {x}^{2} - 14 x + 49 = {\left(x - 7\right)}^{2}$

If we write the original $42 = 49 - 7$:
$\to {\left(x - 7\right)}^{2} - 7 = 0 \to {\left(x - 7\right)}^{2} = 7$

So: $x - 7 = \sqrt{7} \to x = 7 + \sqrt{7}$
Or: $x - 7 = - \sqrt{7} \to x = 7 - \sqrt{7}$

Often written as ${x}_{1 , 2} = 7 \pm \sqrt{7}$