How do you solve the quadratic 9y^2+6y-8=0 using any method?

Feb 13, 2017

$y = \frac{2}{3} \mathmr{and} y = - \frac{4}{3}$

Explanation:

$9 {y}^{2} + 6 y - 8 = 0$

$\therefore \left(3 y - 2\right) \left(3 y + 4\right)$

$\therefore 3 y - 2 = 0 \mathmr{and} 3 y + 4 = 0$

$\therefore 3 y = 2 \mathmr{and} 3 y = - 4$

$\therefore y = \frac{3}{2} \mathmr{and} y = - \frac{4}{3}$

substitute $y = \frac{3}{2}$

$\therefore 9 {\left(\frac{2}{3}\right)}^{2} + 6 \left(\frac{2}{3}\right) - 8 = 0$

$\therefore 9 \left(\frac{4}{9}\right) + \frac{12}{3} - 8 = 0$

$\therefore 4 + 4 - 8 = 0$

$\therefore 8 - 8 = 0$

substitute $y = - \frac{4}{3}$

$\therefore 9 {\left(- \frac{4}{3}\right)}^{2} + 6 \left(- \frac{4}{3}\right) - 8 = 0$

$: 9 \left(\frac{16}{9}\right) - \frac{24}{3} - 8 = 0$

$\therefore 16 - 8 - 8 = 0$

$\therefore 8 - 8 = 0$