What are the solutions to the equation -2v^2-v+12=-3v^2+6v ?

The solutions to the equation -2v^2-v+12=-3v^2+6v are v=4,v=3

Find the zeros of -2v^2-v+12=-3v^2+6v

The zeros of -2v^2-v+12=-3v^2+6v are v=4,v=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 2 v^{2} - v + 12 = - 3 v^{2} + 6 v

v = 4, v = 3

Solution steps

- 2 v^{2} - v + 12 = - 3 v^{2} + 6 v

Move

6 v

to the left side

- 2 v^{2} - 7 v + 12 = - 3 v^{2}

Move

3 v^{2}

to the left side

v^{2} - 7 v + 12 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 12 = 1

v_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 1}, v_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 1}

v = \frac{- ( - 7 ) + 1}{2 \cdot 1} : 4

v = \frac{- ( - 7 ) - 1}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

v = 4, v = 3

4 x^{2} - 20 x + 1 = 0

t^{2} - 8 t - 5 = 0

x^{2} + 6.5 x + 10 = 0

x^{2} + 4 x + 18 = 0

x^{2} - 10 x + 25 = 49

What are the solutions to the equation -2v^2-v+12=-3v^2+6v ?
The solutions to the equation -2v^2-v+12=-3v^2+6v are v=4,v=3
Find the zeros of -2v^2-v+12=-3v^2+6v
The zeros of -2v^2-v+12=-3v^2+6v are v=4,v=3