How do you solve 4x^2+6x=12 by completing the square?

Mar 21, 2017

The solutions are $S = \left\{- 2.637 , 1.137\right\}$

Explanation:

Before completing the square, we divide by $2$

$4 {x}^{2} + 6 x = 12$

$2 {x}^{2} + 3 x = 6$

$2 \left({x}^{2} + \frac{3}{2} x\right) = 6$

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

Now, we complete the square

$\left({x}^{2} + \frac{3}{2} x + \frac{9}{16}\right) = 3 + \frac{9}{16}$

${\left(x + \frac{3}{4}\right)}^{2} = \frac{57}{16}$

$x + \frac{3}{4} = \pm \frac{\sqrt{57}}{16}$

$x = - \frac{3}{4} \pm \frac{\sqrt{57}}{16}$

$x = \frac{1}{4} \left(- 3 \pm \sqrt{57}\right)$

Therefore,

${x}_{1} = \frac{1}{4} \left(- 3 - \sqrt{57}\right) = - 2.637$

${x}_{2} = \frac{1}{4} \left(- 3 + \sqrt{57}\right) = 1.137$