What are the solutions to the equation w^2+2w-24=0 ?

The solutions to the equation w^2+2w-24=0 are w=4,w=-6

Find the zeros of w^2+2w-24=0

The zeros of w^2+2w-24=0 are w=4,w=-6

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w^{2} + 2 w - 24 = 0

w = 4, w = - 6

Solution steps

w^{2} + 2 w - 24 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 24) = 10

w_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 2 + 10}{2 \cdot 1}, w_{2} = \frac{- 2 - 10}{2 \cdot 1}

w = \frac{- 2 + 10}{2 \cdot 1} : 4

w = \frac{- 2 - 10}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

w = 4, w = - 6

(2 x + 1)^{2} = 9

10 \cdot \frac{1}{5}

5 (0) + 7 (2) \geq - 3

- x^{2} - 8 = - 33

1 0^{2} = r^{2} + r^{2} - 2 \cdot r \cdot r \cdot \frac{1 + 5}{4}

What are the solutions to the equation w^2+2w-24=0 ?
The solutions to the equation w^2+2w-24=0 are w=4,w=-6
Find the zeros of w^2+2w-24=0
The zeros of w^2+2w-24=0 are w=4,w=-6