Question

Ayuda..!!!

3n^2-n=102

a^2+a=132

Answer 1

$3n^2-n=102\\ </p><p>3n^2-n-102=102-102\\ </p><p>3n^2-n-102=0\\ </p><p>\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ </p><p>x1=\frac{-\left(-1\right)+\sqrt{\left(-1\right)^2-4\cdot \:3\left(-102\right)}}{2\cdot \:3}\\ </p><p>x2=\frac{-\left(-1\right)-\sqrt{\left(-1\right)^2-4\cdot \:3\left(-102\right)}}{2\cdot \:3}\\ \\ </p><p> n=6,\:n=-\frac{17}{3}$

Answer 2

$3 {n}^{2} - n = 102$

si n pertenece a los reales, entonces:
${3n}^{2} - n - 10 2 = 0 \\ \\3 ( {n}^{2} - \frac{n}{3} - 34) = 0 \\ \\ {n}^{2} - \frac{n}{3} - 34 = 0 \\ \\ {(n - \frac{1}{6}) }^{2} - 34 - \frac{1}{36} = 0 \\ \\ (n - \frac{1}{6} {)}^{2} - \frac{1225}{36} = 0 \\ \\ ( {n - \frac{1}{6}) }^{2} = \frac{1225}{36} \\ \\ \sqrt{ {(n - \frac{1}{6} )}^{2} } = \sqrt{ \frac{1225}{36} } \\ \\ \\ |n - \frac{1}{6} | = \frac{35}{6}$

entonces por definición:
$n - \frac{1}{6} = \frac{35}{6} \\ \\ n = \frac{35}{6} + \frac{1}{6} \\ \\ n = \frac{36}{6} = 6$

o

$n - \frac{1}{6} = - \frac{35}{6} \\ \\ n = - \frac{35}{6} + \frac{1}{6} \\ \\ n = \frac{ - 34}{6} \\ \\ n = - \frac{17}{3}$