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

Question

1789 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-26T12:40:40

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

$(2(x+3)^2)/3-4/9=1/3$

$=>(2(x+3)^2)/3=1/3+4/9$

$=>(2(x+3)^2)/3=1/3 xx3/3+4/9$

$=>(2(x+3)^2)/3=3/9+4/9$

$=>2/3xx(x+3)^2=7/9$

$=>(x+3)^2=7/9xx 3/2$

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

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

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

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

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

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

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

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=1/3(2(x+3)^2)/3-4/9=1/3?