What are the solutions to the equation 400n=n^2+10000 ?

The solutions to the equation 400n=n^2+10000 are n=100(2+sqrt(3)),n=100(2-sqrt(3))

Find the zeros of 400n=n^2+10000

The zeros of 400n=n^2+10000 are n=100(2+sqrt(3)),n=100(2-sqrt(3))

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400 n = n^{2} + 10000

n = 100 (2 + 3), n = 100 (2 - 3)

Decimal Notation

n = 373.20508 \dots, n = 26.79491 \dots

Solution steps

400 n = n^{2} + 10000

Switch sides

n^{2} + 10000 = 400 n

Move

400 n

to the left side

n^{2} + 10000 - 400 n = 0

Write in the standard form

a x^{2} + b x + c = 0

n^{2} - 400 n + 10000 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 400 ) \pm ( - 400 ) ^{2} - 4 \cdot 1 \cdot 10000}{2 \cdot 1}

(- 400)^{2} - 4 \cdot 1 \cdot 10000 = 2003

n_{1, 2} = \frac{- ( - 400 ) \pm 200 3}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 400 ) + 200 3}{2 \cdot 1}, n_{2} = \frac{- ( - 400 ) - 200 3}{2 \cdot 1}

n = \frac{- ( - 400 ) + 200 3}{2 \cdot 1} : 100 (2 + 3)

n = \frac{- ( - 400 ) - 200 3}{2 \cdot 1} : 100 (2 - 3)

The solutions to the quadratic equation are:

n = 100 (2 + 3), n = 100 (2 - 3)

p^{2} - 6 p + 5 = 0

3 x^{2} + 25 = 0

- x^{2} + 10 x - 28

0 = x^{2} + 25

r^{2} - 12 r + 32 = 0

What are the solutions to the equation 400n=n^2+10000 ?
The solutions to the equation 400n=n^2+10000 are n=100(2+sqrt(3)),n=100(2-sqrt(3))
Find the zeros of 400n=n^2+10000
The zeros of 400n=n^2+10000 are n=100(2+sqrt(3)),n=100(2-sqrt(3))