What are the solutions to the equation 3q(q-3)=12 ?

The solutions to the equation 3q(q-3)=12 are q=4,q=-1

Find the zeros of 3q(q-3)=12

The zeros of 3q(q-3)=12 are q=4,q=-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 q (q - 3) = 12

q = 4, q = - 1

Solution steps

3 q (q - 3) = 12

Expand

3 q (q - 3) : 3 q^{2} - 9 q

3 q^{2} - 9 q = 12

Move

12

to the left side

3 q^{2} - 9 q - 12 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 3 ( - 12 )}{2 \cdot 3}

(- 9)^{2} - 4 \cdot 3 (- 12) = 15

q_{1, 2} = \frac{- ( - 9 ) \pm 15}{2 \cdot 3}

Separate the solutions

q_{1} = \frac{- ( - 9 ) + 15}{2 \cdot 3}, q_{2} = \frac{- ( - 9 ) - 15}{2 \cdot 3}

q = \frac{- ( - 9 ) + 15}{2 \cdot 3} : 4

q = \frac{- ( - 9 ) - 15}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

q = 4, q = - 1

x^{2} + 3 x + \frac{13}{4}

2 x^{2} = 25

2 x^{2} = 48

2 x^{2} = 80

7 x^{2} + 3 x - 4 = 0

What are the solutions to the equation 3q(q-3)=12 ?
The solutions to the equation 3q(q-3)=12 are q=4,q=-1
Find the zeros of 3q(q-3)=12
The zeros of 3q(q-3)=12 are q=4,q=-1