Answer
$x=\frac{1}{3}$ and $x=-\frac{7}{3}$
Work Step by Step
The given equation is
$\Rightarrow 9(x+1)^2=16$
Divide each side by $9$.
$\Rightarrow \frac{9(x+1)^2}{9}=\frac{16}{9}$
Simplify.
$\Rightarrow (x+1)^2=\frac{16}{9}$
Take the square root of each side.
$\Rightarrow x+1=\pm\sqrt {\frac{16}{9}}$
Simplify.
$\Rightarrow x+1=\pm\frac{4}{3}$
Subtract $1$ from each side.
$\Rightarrow x+1-1=-1\pm\frac{4}{3}$
Simplify.
$\Rightarrow x=-1\pm\frac{4}{3}$
Two real solutions are
$\Rightarrow x=-1+\frac{4}{3}$ or $x=-1-\frac{4}{3}$
Simplify.
$\Rightarrow x=\frac{1}{3}$ or $x=-\frac{7}{3}$
Hence, the solutions are $x=\frac{1}{3}$ and $x=-\frac{7}{3}$.
