Solve for \( r \). Separate multiple answers with a comma. \[ 225 r^{3}+441 r=-630 r^{2} \]

To solve for \( r \) in the equation \( 225 r^{3}+441 r=-630 r^{2} \), we first rearrange the equation to set it to zero: \[ 225 r^{3} + 630 r^{2} + 441 r = 0 \] Next, we can factor out the common term \( r \): \[ r(225 r^{2} + 630 r + 441) = 0 \] This gives us one solution: \[ r = 0 \] Now we need to solve the quadratic equation \( 225 r^{2} + 630 r + 441 = 0 \). We can use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 225 \), \( b = 630 \), and \( c = 441 \). Calculating the discriminant: \[ b^2 - 4ac = 630^2 - 4 \times 225 \times 441 \] Calculating \( 630^2 \): \[ 630^2 = 396900 \] Calculating \( 4 \times 225 \times 441 \): \[ 4 \times 225 = 900 \] \[ 900 \times 441 = 396900 \] Now, substituting back into the discriminant: \[ b^2 - 4ac = 396900 - 396900 = 0 \] With a discriminant of zero, there is one repeated solution for the quadratic: \[ r = \frac{-630 \pm 0}{2 \times 225} = \frac{-630}{450} = -\frac{14}{10} = -\frac{7}{5} \] Thus, the solutions for \( r \) are: \[ r = 0, -\frac{7}{5} \] Final answer: \[ \boxed{0, -\frac{7}{5}} \]