What are the solutions to the equation 4x^2+x-5=-6x ?

The solutions to the equation 4x^2+x-5=-6x are x=(-7+sqrt(129))/8 ,x=(-7-sqrt(129))/8

Find the zeros of 4x^2+x-5=-6x

The zeros of 4x^2+x-5=-6x are x=(-7+sqrt(129))/8 ,x=(-7-sqrt(129))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 x^{2} + x - 5 = - 6 x

x = \frac{- 7 + 129}{8}, x = \frac{- 7 - 129}{8}

Decimal Notation

x = 0.54472 \dots, x = - 2.29472 \dots

Solution steps

4 x^{2} + x - 5 = - 6 x

Move

6 x

to the left side

4 x^{2} + 7 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

7^{2} - 4 \cdot 4 (- 5) = 129

x_{1, 2} = \frac{- 7 \pm 129}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 7 + 129}{2 \cdot 4}, x_{2} = \frac{- 7 - 129}{2 \cdot 4}

x = \frac{- 7 + 129}{2 \cdot 4} : \frac{- 7 + 129}{8}

x = \frac{- 7 - 129}{2 \cdot 4} : \frac{- 7 - 129}{8}

The solutions to the quadratic equation are:

x = \frac{- 7 + 129}{8}, x = \frac{- 7 - 129}{8}

4 x^{2} + 3 x = x (2 x - 2)

(x - 6) (2 x + 1) = 0

12 x = - 4 x^{2}

x^{2} - 2 a x + a^{2} + b^{2} = 0

x^{2} + 15 x = - 57

What are the solutions to the equation 4x^2+x-5=-6x ?
The solutions to the equation 4x^2+x-5=-6x are x=(-7+sqrt(129))/8 ,x=(-7-sqrt(129))/8
Find the zeros of 4x^2+x-5=-6x
The zeros of 4x^2+x-5=-6x are x=(-7+sqrt(129))/8 ,x=(-7-sqrt(129))/8