What are the solutions to the equation 5x^2+2x-4=0 ?

The solutions to the equation 5x^2+2x-4=0 are x=(-1+sqrt(21))/5 ,x=-(1+sqrt(21))/5

Find the zeros of 5x^2+2x-4=0

The zeros of 5x^2+2x-4=0 are x=(-1+sqrt(21))/5 ,x=-(1+sqrt(21))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{- 1 + 21}{5}, x = - \frac{1 + 21}{5}

Decimal Notation

x = 0.71651 \dots, x = - 1.11651 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 5 (- 4) = 221

x_{1, 2} = \frac{- 2 \pm 2 21}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 2 + 2 21}{2 \cdot 5}, x_{2} = \frac{- 2 - 2 21}{2 \cdot 5}

x = \frac{- 2 + 2 21}{2 \cdot 5} : \frac{- 1 + 21}{5}

x = \frac{- 2 - 2 21}{2 \cdot 5} : - \frac{1 + 21}{5}

The solutions to the quadratic equation are:

x = \frac{- 1 + 21}{5}, x = - \frac{1 + 21}{5}

- 36 + 7 x = 76 - x

6 x^{3} - 24 x^{2} + 6 x

6 x^{2} + 12 x = 90

\frac{2 + 5 i}{3 - i}

\frac{6}{4} - \frac{2}{4}

What are the solutions to the equation 5x^2+2x-4=0 ?
The solutions to the equation 5x^2+2x-4=0 are x=(-1+sqrt(21))/5 ,x=-(1+sqrt(21))/5
Find the zeros of 5x^2+2x-4=0
The zeros of 5x^2+2x-4=0 are x=(-1+sqrt(21))/5 ,x=-(1+sqrt(21))/5