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

Question

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

UpStudy AI · Accepted Answer

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 $$ x = -3 $$.

6) y = -4, y = -12 8) y = -14/3, y = -14 10) x = 1/3, x = -3

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

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