What are the solutions to the equation (3p+3)^2-1(3p+3)-20=0 ?

The solutions to the equation (3p+3)^2-1(3p+3)-20=0 are p= 2/3 ,p=-7/3

Find the zeros of (3p+3)^2-1(3p+3)-20=0

The zeros of (3p+3)^2-1(3p+3)-20=0 are p= 2/3 ,p=-7/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(3 p + 3)^{2} - 1 (3 p + 3) - 20 = 0

p = \frac{2}{3}, p = - \frac{7}{3}

Decimal Notation

p = 0.66666 \dots, p = - 2.33333 \dots

Solution steps

(3 p + 3)^{2} - 1 \cdot (3 p + 3) - 20 = 0

Expand

(3 p + 3)^{2} - 1 \cdot (3 p + 3) - 20 : 9 p^{2} + 15 p - 14

9 p^{2} + 15 p - 14 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 9 ( - 14 )}{2 \cdot 9}

1 5^{2} - 4 \cdot 9 (- 14) = 27

p_{1, 2} = \frac{- 15 \pm 27}{2 \cdot 9}

Separate the solutions

p_{1} = \frac{- 15 + 27}{2 \cdot 9}, p_{2} = \frac{- 15 - 27}{2 \cdot 9}

p = \frac{- 15 + 27}{2 \cdot 9} : \frac{2}{3}

p = \frac{- 15 - 27}{2 \cdot 9} : - \frac{7}{3}

The solutions to the quadratic equation are:

p = \frac{2}{3}, p = - \frac{7}{3}

6 x^{2} + 11 x = 72

0 = - 16 t^{2} + 108 t + 520

\frac{5}{4 i}

\frac{2}{15} + \frac{5}{9}

25 rs + 25 r - 19 s - 19

What are the solutions to the equation (3p+3)^2-1(3p+3)-20=0 ?
The solutions to the equation (3p+3)^2-1(3p+3)-20=0 are p= 2/3 ,p=-7/3
Find the zeros of (3p+3)^2-1(3p+3)-20=0
The zeros of (3p+3)^2-1(3p+3)-20=0 are p= 2/3 ,p=-7/3