What are the solutions to the equation 4v^2+7v=1 ?

The solutions to the equation 4v^2+7v=1 are v=(-7+sqrt(65))/8 ,v=(-7-sqrt(65))/8

Find the zeros of 4v^2+7v=1

The zeros of 4v^2+7v=1 are v=(-7+sqrt(65))/8 ,v=(-7-sqrt(65))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 v^{2} + 7 v = 1

v = \frac{- 7 + 65}{8}, v = \frac{- 7 - 65}{8}

Decimal Notation

v = 0.13278 \dots, v = - 1.88278 \dots

Solution steps

4 v^{2} + 7 v = 1

Move

1

to the left side

4 v^{2} + 7 v - 1 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 4 (- 1) = 65

v_{1, 2} = \frac{- 7 \pm 65}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- 7 + 65}{2 \cdot 4}, v_{2} = \frac{- 7 - 65}{2 \cdot 4}

v = \frac{- 7 + 65}{2 \cdot 4} : \frac{- 7 + 65}{8}

v = \frac{- 7 - 65}{2 \cdot 4} : \frac{- 7 - 65}{8}

The solutions to the quadratic equation are:

v = \frac{- 7 + 65}{8}, v = \frac{- 7 - 65}{8}

4 x^{2} + 4 x = 11

- 2 h - 1 = - 1 - 2 h

4 - 49 x^{2}

x^{2} - 10 x = 21

y \cdot (x^{2} + 3) - 2 \cdot (x^{2} + 3)

What are the solutions to the equation 4v^2+7v=1 ?
The solutions to the equation 4v^2+7v=1 are v=(-7+sqrt(65))/8 ,v=(-7-sqrt(65))/8
Find the zeros of 4v^2+7v=1
The zeros of 4v^2+7v=1 are v=(-7+sqrt(65))/8 ,v=(-7-sqrt(65))/8