What are the solutions to the equation 400=n((2(-5)+4(n-1))/2) ?

The solutions to the equation 400=n((2(-5)+4(n-1))/2) are n=16,n=-25/2

Find the zeros of 400=n((2(-5)+4(n-1))/2)

The zeros of 400=n((2(-5)+4(n-1))/2) are n=16,n=-25/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

400 = n (\frac{2 ( - 5 ) + 4 ( n - 1 )}{2})

n = 16, n = - \frac{25}{2}

Decimal Notation

n = 16, n = - 12.5

Solution steps

400 = n (\frac{2 ( - 5 ) + 4 ( n - 1 )}{2})

Expand

n (\frac{2 ( - 5 ) + 4 ( n - 1 )}{2}) : 2 n^{2} - 7 n

400 = 2 n^{2} - 7 n

Switch sides

2 n^{2} - 7 n = 400

Move

400

to the left side

2 n^{2} - 7 n - 400 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 400 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 400) = 57

n_{1, 2} = \frac{- ( - 7 ) \pm 57}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- ( - 7 ) + 57}{2 \cdot 2}, n_{2} = \frac{- ( - 7 ) - 57}{2 \cdot 2}

n = \frac{- ( - 7 ) + 57}{2 \cdot 2} : 16

n = \frac{- ( - 7 ) - 57}{2 \cdot 2} : - \frac{25}{2}

The solutions to the quadratic equation are:

n = 16, n = - \frac{25}{2}

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

0.12 \cdot 1 0^{- 5} x^{2} + 0.026 x + 100 = 0

(x + \frac{b}{3})^{2} - 2 - \frac{b ^{2}}{9} = 0

2 x^{2} - x - 2

(2 x + 3)^{2} = 4

What are the solutions to the equation 400=n((2(-5)+4(n-1))/2) ?
The solutions to the equation 400=n((2(-5)+4(n-1))/2) are n=16,n=-25/2
Find the zeros of 400=n((2(-5)+4(n-1))/2)
The zeros of 400=n((2(-5)+4(n-1))/2) are n=16,n=-25/2