What are the solutions to the equation n^2=56+n ?

The solutions to the equation n^2=56+n are n=8,n=-7

Find the zeros of n^2=56+n

The zeros of n^2=56+n are n=8,n=-7

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n^{2} = 56 + n

n = 8, n = - 7

Solution steps

n^{2} = 56 + n

Move

n

to the left side

n^{2} - n = 56

Move

56

to the left side

n^{2} - n - 56 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 56 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 56) = 15

n_{1, 2} = \frac{- ( - 1 ) \pm 15}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 15}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 15}{2 \cdot 1}

n = \frac{- ( - 1 ) + 15}{2 \cdot 1} : 8

n = \frac{- ( - 1 ) - 15}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

n = 8, n = - 7

h^{2} = 10 0^{2} + 10 0^{2}

8 - x^{2} = 4 x + 12

5 x^{2} + 7 x^{2} - 25 5^{2} = 0

2 x^{2} - x - 60 = 0

y, y^{2} = \frac{x ^{3}}{2 - x}

What are the solutions to the equation n^2=56+n ?
The solutions to the equation n^2=56+n are n=8,n=-7
Find the zeros of n^2=56+n
The zeros of n^2=56+n are n=8,n=-7