What are the solutions to the equation n^2-n-240=0 ?

The solutions to the equation n^2-n-240=0 are n=16,n=-15

Find the zeros of n^2-n-240=0

The zeros of n^2-n-240=0 are n=16,n=-15

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n^{2} - n - 240 = 0

n = 16, n = - 15

Solution steps

n^{2} - n - 240 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 240 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 240) = 31

n_{1, 2} = \frac{- ( - 1 ) \pm 31}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 31}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 31}{2 \cdot 1}

n = \frac{- ( - 1 ) + 31}{2 \cdot 1} : 16

n = \frac{- ( - 1 ) - 31}{2 \cdot 1} : - 15

The solutions to the quadratic equation are:

n = 16, n = - 15

z^{2} + 14 z + 45 = 0

x^{2} - 2 x

x, x^{2} + 7 x - 18 = 0

2 x^{2} - x - 55 = 0

2 x^{2} - 5 x - 13 = 0

What are the solutions to the equation n^2-n-240=0 ?
The solutions to the equation n^2-n-240=0 are n=16,n=-15
Find the zeros of n^2-n-240=0
The zeros of n^2-n-240=0 are n=16,n=-15