What are the solutions to the equation w^2+4w-12=0 ?

The solutions to the equation w^2+4w-12=0 are w=2,w=-6

Find the zeros of w^2+4w-12=0

The zeros of w^2+4w-12=0 are w=2,w=-6

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w^{2} + 4 w - 12 = 0

w = 2, w = - 6

Solution steps

w^{2} + 4 w - 12 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 12) = 8

w_{1, 2} = \frac{- 4 \pm 8}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 4 + 8}{2 \cdot 1}, w_{2} = \frac{- 4 - 8}{2 \cdot 1}

w = \frac{- 4 + 8}{2 \cdot 1} : 2

w = \frac{- 4 - 8}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

w = 2, w = - 6

k^{2} + 3 k = 0

- x^{2} - 100 = 0

49 k^{2} + 7 k - 11 = 1 - 5 k + 4 k^{2}

x^{2} + 10 x + 24 = 0

8^{2} + 1 7^{2} = c^{2}

What are the solutions to the equation w^2+4w-12=0 ?
The solutions to the equation w^2+4w-12=0 are w=2,w=-6
Find the zeros of w^2+4w-12=0
The zeros of w^2+4w-12=0 are w=2,w=-6