# How do you solve the equation (2(x+3)^2)/3-4/9=1/3?

Oct 26, 2017

$x = - 4.0801$
Or
$x = - 1.9198$

#### Explanation:

$\frac{2 {\left(x + 3\right)}^{2}}{3} - \frac{4}{9} = \frac{1}{3}$

$\implies \frac{2 {\left(x + 3\right)}^{2}}{3} = \frac{1}{3} + \frac{4}{9}$

$\implies \frac{2 {\left(x + 3\right)}^{2}}{3} = \frac{1}{3} \times \frac{3}{3} + \frac{4}{9}$

$\implies \frac{2 {\left(x + 3\right)}^{2}}{3} = \frac{3}{9} + \frac{4}{9}$

$\implies \frac{2}{3} \times {\left(x + 3\right)}^{2} = \frac{7}{9}$

$\implies {\left(x + 3\right)}^{2} = \frac{7}{9} \times \frac{3}{2}$

$\implies {\left(x + 3\right)}^{2} = \frac{7}{\cancel{9}} ^ 3 \times {\cancel{3}}^{1} / 2$

$\implies {x}^{2} + 6 x + 9 = \frac{7}{6}$

$\implies {x}^{2} + 6 x + 9 - \frac{7}{6} = 0$

$\implies {x}^{2} + 6 x + \frac{9 \times 6}{6} - \frac{7}{6} = 0$

$\implies {x}^{2} + 6 x + \frac{54 - 7}{6} = 0$

${x}^{2} + 6 x + \frac{47}{6} = 0$

$6 {x}^{2} + 36 x + 47 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Where $a = 6 , b = 36 \mathmr{and} c = 47$
We get the truncated approximate values of $x$ as
$x = - 4.0801$
$x = - 1.9198$