How do you solve the equation $\frac{2 {(x + 3)}^{2}}{3} - \frac{4}{9} = \frac{1}{3}$ $(2(x+3)^2)/3-4/9=1/3$ ?

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Answer 1 · 2017-10-26T12:40:40

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

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

$\Rightarrow \frac{2 {(x + 3)}^{2}}{3} = \frac{1}{3} + \frac{4}{9}$ $=>(2(x+3)^2)/3=1/3+4/9$

$\Rightarrow \frac{2 {(x + 3)}^{2}}{3} = \frac{1}{3} \times \frac{3}{3} + \frac{4}{9}$ $=>(2(x+3)^2)/3=1/3 xx3/3+4/9$

$\Rightarrow \frac{2 {(x + 3)}^{2}}{3} = \frac{3}{9} + \frac{4}{9}$ $=>(2(x+3)^2)/3=3/9+4/9$

$\Rightarrow \frac{2}{3} \times {(x + 3)}^{2} = \frac{7}{9}$ $=>2/3xx(x+3)^2=7/9$

$\Rightarrow {(x + 3)}^{2} = \frac{7}{9} \times \frac{3}{2}$ $=>(x+3)^2=7/9xx 3/2$

$=>(x+3)^2=7/cancel9^3xx cancel3^1/2$

$=>x^2 + 6x + 9 = 7/6$

$=>x^2 + 6x + 9 - 7/6 = 0$

$=>x^2 + 6x + (9xx6)/6 - 7/6 = 0$

$=>x^2 + 6x + (54 - 7)/6 = 0$

$x^2 + 6x + 47/6 = 0$

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

Using quadratic formula :
$x=( -b +-sqrt(b^2 -4ac))/(2a)$

Where $a=6, b=36 and c =47$

We get the truncated approximate values of $x$ as
$x = -4.0801$
Or
$x= -1.9198$

How do you solve the equation (2(x+3)^2)/3-4/9=1/32(x+3)23−49=13(2(x+3)^2)/3-4/9=1/3?