What are the solutions to the equation 1=5x^2+4x ?

The solutions to the equation 1=5x^2+4x are x= 1/5 ,x=-1

Find the zeros of 1=5x^2+4x

The zeros of 1=5x^2+4x are x= 1/5 ,x=-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1 = 5 x^{2} + 4 x

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

Decimal Notation

x = 0.2, x = - 1

Solution steps

1 = 5 x^{2} + 4 x

Switch sides

5 x^{2} + 4 x = 1

Move

1

to the left side

5 x^{2} + 4 x - 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 5 (- 1) = 6

x_{1, 2} = \frac{- 4 \pm 6}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 4 + 6}{2 \cdot 5}, x_{2} = \frac{- 4 - 6}{2 \cdot 5}

x = \frac{- 4 + 6}{2 \cdot 5} : \frac{1}{5}

x = \frac{- 4 - 6}{2 \cdot 5} : - 1

The solutions to the quadratic equation are:

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

x^{2} + 14 x = 12

1 5^{2} + x^{2} = 2 0^{2}

(x - 6)^{2} = 44

7 v^{2} - 3 v = 0

3 x^{2} - 6 = 24

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