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

The solutions to the equation 4n^2+5n-5=0 are n=(-5+sqrt(105))/8 ,n=(-5-sqrt(105))/8

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

The zeros of 4n^2+5n-5=0 are n=(-5+sqrt(105))/8 ,n=(-5-sqrt(105))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

n = \frac{- 5 + 105}{8}, n = \frac{- 5 - 105}{8}

Decimal Notation

n = 0.65586 \dots, n = - 1.90586 \dots

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 4 (- 5) = 105

n_{1, 2} = \frac{- 5 \pm 105}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- 5 + 105}{2 \cdot 4}, n_{2} = \frac{- 5 - 105}{2 \cdot 4}

n = \frac{- 5 + 105}{2 \cdot 4} : \frac{- 5 + 105}{8}

n = \frac{- 5 - 105}{2 \cdot 4} : \frac{- 5 - 105}{8}

The solutions to the quadratic equation are:

n = \frac{- 5 + 105}{8}, n = \frac{- 5 - 105}{8}

\frac{1}{4} h^{2} - \frac{3}{4} h = 0

(\frac{r}{s})^{4}

12 x^{2} - 4 x = 20 + 10 x

(- x - 4) \cdot 2 = 3 \cdot (- 3 x + 1) + 12

2^{n} \cdot 3^{n}

What are the solutions to the equation 4n^2+5n-5=0 ?
The solutions to the equation 4n^2+5n-5=0 are n=(-5+sqrt(105))/8 ,n=(-5-sqrt(105))/8
Find the zeros of 4n^2+5n-5=0
The zeros of 4n^2+5n-5=0 are n=(-5+sqrt(105))/8 ,n=(-5-sqrt(105))/8