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

Question

2059 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-23T11:04:25

$x \approx - 1.13 or x \approx - 3.54$ $x~~ -1.13 or x~~ -3.54$

$- 3 x^{2} - 12 = 14 x or 3 x^{2} + 14 x + 12 = 0 or 3 (x^{2} + \frac{14}{3} x) + 12$ $-3x^2-12=14x or 3x^2+14x+12 =0 or 3(x^2+14/3x)+12$ or

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

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

$3 {(x + \frac{7}{3})}^{2} - \frac{13}{3} = 0 or 3 {(x + \frac{7}{3})}^{2} = \frac{13}{3}$ $3(x+7/3)^2-13/3 =0 or 3(x+7/3)^2=13/3$ or

${(x + \frac{7}{3})}^{2} = \frac{13}{9} or (x + \frac{7}{3}) = \pm \frac{\sqrt{13}}{3}$ $(x+7/3)^2=13/9 or (x+7/3) = +-sqrt(13)/3$ . Either

$x = - \frac{7}{3} + \frac{\sqrt{13}}{3} or x = - \frac{7}{3} - \frac{\sqrt{13}}{3}$ $x= -7/3+sqrt13/3 or x= -7/3-sqrt13/3$ or

$x = \frac{\sqrt{13} - 7}{3} or x = - \frac{\sqrt{13} + 7}{3}$ $x= (sqrt13-7)/3 or x= -(sqrt13+7)/3$ or

$x \approx - 1.13 (2 d p) or x \approx - 3.54 (2 d p)$ $x~~ -1.13(2dp) or x~~ -3.54(2dp)$ [Ans]

How do you solve -3x^2-12=14x−3x2−12=14x-3x^2-12=14x by completing the square?