What are the solutions to the equation n^2+12n-28=0 ?

The solutions to the equation n^2+12n-28=0 are n=2,n=-14

Find the zeros of n^2+12n-28=0

The zeros of n^2+12n-28=0 are n=2,n=-14

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n^{2} + 12 n - 28 = 0

n = 2, n = - 14

Solution steps

n^{2} + 12 n - 28 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 28) = 16

n_{1, 2} = \frac{- 12 \pm 16}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 12 + 16}{2 \cdot 1}, n_{2} = \frac{- 12 - 16}{2 \cdot 1}

n = \frac{- 12 + 16}{2 \cdot 1} : 2

n = \frac{- 12 - 16}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

n = 2, n = - 14

6 = \frac{3}{4} (10 x + 16)

\frac{4 ^{7}}{4}

- 24 = - 3 w + 21

w^{2} = 9 (w - 8)

- 12 t > 36

What are the solutions to the equation n^2+12n-28=0 ?
The solutions to the equation n^2+12n-28=0 are n=2,n=-14
Find the zeros of n^2+12n-28=0
The zeros of n^2+12n-28=0 are n=2,n=-14