What are the solutions to the equation u^2+(4u)/5 =5 ?

The solutions to the equation u^2+(4u)/5 =5 are u=(-2+sqrt(129))/5 ,u=-(2+sqrt(129))/5

Find the zeros of u^2+(4u)/5 =5

The zeros of u^2+(4u)/5 =5 are u=(-2+sqrt(129))/5 ,u=-(2+sqrt(129))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

u^{2} + \frac{4 u}{5} = 5

u = \frac{- 2 + 129}{5}, u = - \frac{2 + 129}{5}

Decimal Notation

u = 1.87156 \dots, u = - 2.67156 \dots

Solution steps

u^{2} + \frac{4 u}{5} = 5

Multiply both sides by

5

5 u^{2} + 4 u = 25

Move

25

to the left side

5 u^{2} + 4 u - 25 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 5 ( - 25 )}{2 \cdot 5}

4^{2} - 4 \cdot 5 (- 25) = 2129

u_{1, 2} = \frac{- 4 \pm 2 129}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 4 + 2 129}{2 \cdot 5}, u_{2} = \frac{- 4 - 2 129}{2 \cdot 5}

u = \frac{- 4 + 2 129}{2 \cdot 5} : \frac{- 2 + 129}{5}

u = \frac{- 4 - 2 129}{2 \cdot 5} : - \frac{2 + 129}{5}

The solutions to the quadratic equation are:

u = \frac{- 2 + 129}{5}, u = - \frac{2 + 129}{5}

\frac{10}{x + 4} > 3 - x

\frac{1}{3} (2)^{3}

x^{3} x^{7}

\frac{k}{3} 2 + k - 5 + k + \frac{k}{3} = 139

- 8 x - 3 \geq - 4 x + 5

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