What are the solutions to the equation 0=5x^2+2x-12 ?

The solutions to the equation 0=5x^2+2x-12 are x=(-1+sqrt(61))/5 ,x=-(1+sqrt(61))/5

Find the zeros of 0=5x^2+2x-12

The zeros of 0=5x^2+2x-12 are x=(-1+sqrt(61))/5 ,x=-(1+sqrt(61))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = 5 x^{2} + 2 x - 12

x = \frac{- 1 + 61}{5}, x = - \frac{1 + 61}{5}

Decimal Notation

x = 1.36204 \dots, x = - 1.76204 \dots

Solution steps

0 = 5 x^{2} + 2 x - 12

Switch sides

5 x^{2} + 2 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 12 )}{2 \cdot 5}

2^{2} - 4 \cdot 5 (- 12) = 261

x_{1, 2} = \frac{- 2 \pm 2 61}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 2 + 2 61}{2 \cdot 5}, x_{2} = \frac{- 2 - 2 61}{2 \cdot 5}

x = \frac{- 2 + 2 61}{2 \cdot 5} : \frac{- 1 + 61}{5}

x = \frac{- 2 - 2 61}{2 \cdot 5} : - \frac{1 + 61}{5}

The solutions to the quadratic equation are:

x = \frac{- 1 + 61}{5}, x = - \frac{1 + 61}{5}

3 t^{2} - 48 = 0

5 u^{2} + 15 u = 0

\frac{x ^{2}}{4} = 2 x - 10

12 x^{2} + 48 x - 4 = 0

9 b^{2} - 16 = 0

What are the solutions to the equation 0=5x^2+2x-12 ?
The solutions to the equation 0=5x^2+2x-12 are x=(-1+sqrt(61))/5 ,x=-(1+sqrt(61))/5
Find the zeros of 0=5x^2+2x-12
The zeros of 0=5x^2+2x-12 are x=(-1+sqrt(61))/5 ,x=-(1+sqrt(61))/5