# How do you solve -3x^2-12=14x by completing the square?

Jun 23, 2017

#### Answer:

$x \approx - 1.13 \mathmr{and} x \approx - 3.54$

#### Explanation:

$- 3 {x}^{2} - 12 = 14 x \mathmr{and} 3 {x}^{2} + 14 x + 12 = 0 \mathmr{and} 3 \left({x}^{2} + \frac{14}{3} x\right) + 12$ or

$3 \left({x}^{2} + \frac{14}{3} x + {\left(\frac{7}{3}\right)}^{2}\right) - \left(3 \cdot \frac{49}{9}\right) + 12 = 0$

$3 {\left(x + \frac{7}{3}\right)}^{2} - \frac{49}{3} + 12 = 0$ or

$3 {\left(x + \frac{7}{3}\right)}^{2} - \frac{13}{3} = 0 \mathmr{and} 3 {\left(x + \frac{7}{3}\right)}^{2} = \frac{13}{3}$ or

${\left(x + \frac{7}{3}\right)}^{2} = \frac{13}{9} \mathmr{and} \left(x + \frac{7}{3}\right) = \pm \frac{\sqrt{13}}{3}$. Either

$x = - \frac{7}{3} + \frac{\sqrt{13}}{3} \mathmr{and} x = - \frac{7}{3} - \frac{\sqrt{13}}{3}$ or

$x = \frac{\sqrt{13} - 7}{3} \mathmr{and} x = - \frac{\sqrt{13} + 7}{3}$ or

$x \approx - 1.13 \left(2 \mathrm{dp}\right) \mathmr{and} x \approx - 3.54 \left(2 \mathrm{dp}\right)$ [Ans]