Solve the equation
-$3(p+4)^{2}-10(p+4)+3=0$

A) $\left\{-\frac{11}{3},-1\right\}$

B) $\left\{\frac{1}{3}, 3\right\}$

C) $\{-3,-3\}$

D) $\left\{-\frac{13}{3},-7\right\}$

Question

Solve the equation
-$3(p+4)^{2}-10(p+4)+3=0$

A) $\left\{-\frac{11}{3},-1\right\}$

B) $\left\{\frac{1}{3}, 3\right\}$

C) $\{-3,-3\}$

D) $\left\{-\frac{13}{3},-7\right\}$

Quizplus · Accepted Answer

Let $x = p + 4$. The equation becomes $3x^2 - 10x + 3 = 0$. Factoring or using the quadratic formula, we find $x = \frac{10 \pm \sqrt{(-10)^2 - 4(3)(3)}}{2(3)}$, which simplifies to $x = \frac{10 \pm \sqrt{64}}{6}$, giving $x = \frac{10 \pm 8}{6}$. Thus, $x = 3$ or $x = \frac{1}{3}$. Substituting back for $p$, we get $p + 4 = 3$ or $p + 4 = \frac{1}{3}$, leading to $p = -1$ or $p = -\frac{11}{3}$.

Solve the Equation - 3(p+4)2−10(p+4)+3=03(p+4)^{2}-10(p+4)+3=03(p+4)2−10(p+4)+3=0 A) {−113,−1}\left\{-\frac{11}{3},-1\right\}{−311​,−1} B) {13,3}\left\{\frac{1}{3}, 3\right\}{31​,3} C) {−3,−3}\{-3,-3\}{−3,−3} D) {−133,−7}\left\{-\frac{13}{3},-7\right\}{−313​,−7}

Solve the Equation
- $3(p+4)^{2}-10(p+4)+3=0$ A) $\left\{-\frac{11}{3},-1\right\}$ B) $\left\{\frac{1}{3}, 3\right\}$ C) $\{-3,-3\}$ D) $\left\{-\frac{13}{3},-7\right\}$