How do you solve $3/4(2x+3)^2-9=0$ ?

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Answer 1 · 2016-04-30T16:58:05

$x=(-3+-2sqrt3)/2$

$3/4(2x+3)^2-9=0$
$=>3/4(2x+3)^2=9$
$=>(2x+3)^2=9xx4/3=12$
$=>(2x+3)=+-sqrt(3*2^2) =+-2sqrt3$
$=>2x=(-3+-2sqrt3)$
$=>:.x=(-3+-2sqrt3)/2$

How do you solve 3/4(2x+3)^2-9=0 3/4(2x+3)^2-9=0 ?