\( \begin{aligned} \text { 6) } & -3 y^{2}=48 y+144 \\ \text { 8) } & 3 y^{2}+19 y+225=29-9 y \\ \text { 10) } & (2 x+1)^{2}=(2-x)^{2}\end{aligned} \)

Alright, I need to solve these three quadratic equations. Let's tackle them one by one. **Problem 6:** \[ -3y^{2} = 48y + 144 \] First, I'll rearrange the equation to standard quadratic form: \[ -3y^{2} - 48y - 144 = 0 \] To simplify, I can divide the entire equation by -3: \[ y^{2} + 16y + 48 = 0 \] Now, I'll attempt to factor this quadratic equation. I'm looking for two numbers that multiply to 48 and add up to 16. The numbers 4 and 12 fit this requirement since 4 * 12 = 48 and 4 + 12 = 16. \[ (y + 4)(y + 12) = 0 \] Setting each factor equal to zero gives the solutions: \[ y + 4 = 0 \Rightarrow y = -4 \] \[ y + 12 = 0 \Rightarrow y = -12 \] So, the solutions are \( y = -4 \) and \( y = -12 \). **Problem 8:** \[ 3y^{2} + 19y + 225 = 29 - 9y \] First, I'll bring all terms to one side to set the equation to zero: \[ 3y^{2} + 19y + 225 - 29 + 9y = 0 \] Simplifying: \[ 3y^{2} + 28y + 196 = 0 \] Now, I'll attempt to factor this quadratic equation. I'm looking for two numbers that multiply to \( 3 \times 196 = 588 \) and add up to 28. The numbers 14 and 14 fit this requirement since 14 * 14 = 196 and 14 + 14 = 28. \[ (3y + 14)(y + 14) = 0 \] Setting each factor equal to zero gives the solutions: \[ 3y + 14 = 0 \Rightarrow y = -\frac{14}{3} \] \[ y + 14 = 0 \Rightarrow y = -14 \] So, the solutions are \( y = -\frac{14}{3} \) and \( y = -14 \). **Problem 10:** \[ (2x + 1)^{2} = (2 - x)^{2} \] First, I'll expand both sides: \[ 4x^{2} + 4x + 1 = 4 - 4x + x^{2} \] Now, I'll bring all terms to one side to set the equation to zero: \[ 4x^{2} + 4x + 1 - 4 + 4x - x^{2} = 0 \] Simplifying: \[ 3x^{2} + 8x - 3 = 0 \] Now, I'll attempt to factor this quadratic equation. I'm looking for two numbers that multiply to \( 3 \times -3 = -9 \) and add up to 8. The numbers 9 and -1 fit this requirement since 9 * -1 = -9 and 9 + (-1) = 8. \[ (3x - 1)(x + 3) = 0 \] Setting each factor equal to zero gives the solutions: \[ 3x - 1 = 0 \Rightarrow x = \frac{1}{3} \] \[ x + 3 = 0 \Rightarrow x = -3 \] So, the solutions are \( x = \frac{1}{3} \) and \( x = -3 \).