What are the solutions to the equation v^2-7v-5=-6v ?

The solutions to the equation v^2-7v-5=-6v are v=(1+sqrt(21))/2 ,v=(1-sqrt(21))/2

Find the zeros of v^2-7v-5=-6v

The zeros of v^2-7v-5=-6v are v=(1+sqrt(21))/2 ,v=(1-sqrt(21))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

v^{2} - 7 v - 5 = - 6 v

v = \frac{1 + 21}{2}, v = \frac{1 - 21}{2}

Solution steps

v^{2} - 7 v - 5 = - 6 v

Move

6 v

to the left side

v^{2} - v - 5 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 5) = 21

v_{1, 2} = \frac{- ( - 1 ) \pm 21}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 21}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 21}{2 \cdot 1}

v = \frac{- ( - 1 ) + 21}{2 \cdot 1} : \frac{1 + 21}{2}

v = \frac{- ( - 1 ) - 21}{2 \cdot 1} : \frac{1 - 21}{2}

The solutions to the quadratic equation are:

v = \frac{1 + 21}{2}, v = \frac{1 - 21}{2}

\frac{6 !}{3 ! \cdot 3 !}

6 x < - 30

\frac{- 4}{2 x - 1}

\frac{3}{4} - \frac{1}{6} + \frac{3}{8}

2 x - \frac{x}{2} = 54

What are the solutions to the equation v^2-7v-5=-6v ?
The solutions to the equation v^2-7v-5=-6v are v=(1+sqrt(21))/2 ,v=(1-sqrt(21))/2
Find the zeros of v^2-7v-5=-6v
The zeros of v^2-7v-5=-6v are v=(1+sqrt(21))/2 ,v=(1-sqrt(21))/2