What are the solutions to the equation n(n-24)=-128 ?

The solutions to the equation n(n-24)=-128 are n=16,n=8

Find the zeros of n(n-24)=-128

The zeros of n(n-24)=-128 are n=16,n=8

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n (n - 24) = - 128

n = 16, n = 8

Solution steps

n (n - 24) = - 128

Expand

n (n - 24) : n^{2} - 24 n

n^{2} - 24 n = - 128

Move

128

to the left side

n^{2} - 24 n + 128 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 1 \cdot 128}{2 \cdot 1}

(- 24)^{2} - 4 \cdot 1 \cdot 128 = 8

n_{1, 2} = \frac{- ( - 24 ) \pm 8}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 24 ) + 8}{2 \cdot 1}, n_{2} = \frac{- ( - 24 ) - 8}{2 \cdot 1}

n = \frac{- ( - 24 ) + 8}{2 \cdot 1} : 16

n = \frac{- ( - 24 ) - 8}{2 \cdot 1} : 8

The solutions to the quadratic equation are:

n = 16, n = 8

- 16 x^{2} + 144 x = 0

9 x^{2} = 30 x - 28

x (x + 1) = 38 + 6

a, 0 = a^{2} - ab - b^{2}

x^{2} = 25

What are the solutions to the equation n(n-24)=-128 ?
The solutions to the equation n(n-24)=-128 are n=16,n=8
Find the zeros of n(n-24)=-128
The zeros of n(n-24)=-128 are n=16,n=8