What are the solutions to the equation 2n^2+n-2035=0 ?

The solutions to the equation 2n^2+n-2035=0 are n=(-1+9sqrt(201))/4 ,n=(-1-9sqrt(201))/4

Find the zeros of 2n^2+n-2035=0

The zeros of 2n^2+n-2035=0 are n=(-1+9sqrt(201))/4 ,n=(-1-9sqrt(201))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 n^{2} + n - 2035 = 0

n = \frac{- 1 + 9 201}{4}, n = \frac{- 1 - 9 201}{4}

Decimal Notation

n = 31.64925 \dots, n = - 32.14925 \dots

Solution steps

2 n^{2} + n - 2035 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 2035 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 2035) = 9201

n_{1, 2} = \frac{- 1 \pm 9 201}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- 1 + 9 201}{2 \cdot 2}, n_{2} = \frac{- 1 - 9 201}{2 \cdot 2}

n = \frac{- 1 + 9 201}{2 \cdot 2} : \frac{- 1 + 9 201}{4}

n = \frac{- 1 - 9 201}{2 \cdot 2} : \frac{- 1 - 9 201}{4}

The solutions to the quadratic equation are:

n = \frac{- 1 + 9 201}{4}, n = \frac{- 1 - 9 201}{4}

x, 4 x^{2} + y^{2} - 4 z^{2} - 4 = 0

x^{2} - 6 x - 27 = 0

(2 x + 3)^{2} - 108 = 0

x, 2 (m x - 1)^{2} = m (2 x - 1)^{2}

x^{2} + 14 x + 49 = 7

What are the solutions to the equation 2n^2+n-2035=0 ?
The solutions to the equation 2n^2+n-2035=0 are n=(-1+9sqrt(201))/4 ,n=(-1-9sqrt(201))/4
Find the zeros of 2n^2+n-2035=0
The zeros of 2n^2+n-2035=0 are n=(-1+9sqrt(201))/4 ,n=(-1-9sqrt(201))/4