Solve low as \[ (3 x+10)^{2}-81=0 \] NSTRUCTION: Gike yeur answers as simplilied fractions. Answer: \[ \begin{array}{l} z=0 \\ z= \end{array} \]

To solve the equation \((3x + 10)^{2} - 81 = 0\), we can start by isolating the square term: \[ (3x + 10)^{2} = 81 \] Next, we take the square root of both sides, remembering to consider both the positive and negative roots: \[ 3x + 10 = 9 \quad \text{or} \quad 3x + 10 = -9 \] Now, let's solve each equation separately. For the first equation: \[ 3x + 10 = 9 \] Subtracting 10 from both sides gives: \[ 3x = 9 - 10 \implies 3x = -1 \] Now dividing by 3: \[ x = -\frac{1}{3} \] For the second equation: \[ 3x + 10 = -9 \] Subtracting 10 from both sides gives: \[ 3x = -9 - 10 \implies 3x = -19 \] Now dividing by 3: \[ x = -\frac{19}{3} \] Putting it all together, the answers are: \[ \begin{array}{l} z = -\frac{1}{3} \\ z = -\frac{19}{3} \end{array} \]