What are the solutions to the equation n^2+n-12=0 ?

The solutions to the equation n^2+n-12=0 are n=3,n=-4

Find the zeros of n^2+n-12=0

The zeros of n^2+n-12=0 are n=3,n=-4

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n^{2} + n - 12 = 0

n = 3, n = - 4

Solution steps

n^{2} + n - 12 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

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

Separate the solutions

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

n = \frac{- 1 + 7}{2 \cdot 1} : 3

n = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

n = 3, n = - 4

x - 6 = - 3 x^{2}

t^{2} + 22 t - 5 = 0

(x + 4)^{2} = 12

2 x^{2} - \frac{7}{3} x + \frac{14}{4} = 0

12 x^{2} - 4 x = 10 x^{2} - 2

What are the solutions to the equation n^2+n-12=0 ?
The solutions to the equation n^2+n-12=0 are n=3,n=-4
Find the zeros of n^2+n-12=0
The zeros of n^2+n-12=0 are n=3,n=-4