What are the solutions to the equation 5x^2+9x=4 ?

The solutions to the equation 5x^2+9x=4 are x=(-9+sqrt(161))/(10),x=(-9-sqrt(161))/(10)

Find the zeros of 5x^2+9x=4

The zeros of 5x^2+9x=4 are x=(-9+sqrt(161))/(10),x=(-9-sqrt(161))/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 x^{2} + 9 x = 4

x = \frac{- 9 + 161}{10}, x = \frac{- 9 - 161}{10}

Decimal Notation

x = 0.36885 \dots, x = - 2.16885 \dots

Solution steps

5 x^{2} + 9 x = 4

Move

4

to the left side

5 x^{2} + 9 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 5 ( - 4 )}{2 \cdot 5}

9^{2} - 4 \cdot 5 (- 4) = 161

x_{1, 2} = \frac{- 9 \pm 161}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 9 + 161}{2 \cdot 5}, x_{2} = \frac{- 9 - 161}{2 \cdot 5}

x = \frac{- 9 + 161}{2 \cdot 5} : \frac{- 9 + 161}{10}

x = \frac{- 9 - 161}{2 \cdot 5} : \frac{- 9 - 161}{10}

The solutions to the quadratic equation are:

x = \frac{- 9 + 161}{10}, x = \frac{- 9 - 161}{10}

3 x^{3} - 2 x^{2} + 12 x - 8

15 x - 21

x^{(2)} + 2 x - 3 = 0

x (x^{\frac{1}{3}})

2 \cdot \frac{2}{2}

What are the solutions to the equation 5x^2+9x=4 ?
The solutions to the equation 5x^2+9x=4 are x=(-9+sqrt(161))/(10),x=(-9-sqrt(161))/(10)
Find the zeros of 5x^2+9x=4
The zeros of 5x^2+9x=4 are x=(-9+sqrt(161))/(10),x=(-9-sqrt(161))/(10)