What are the solutions to the equation w^2+7w+12=0 ?

The solutions to the equation w^2+7w+12=0 are w=-3,w=-4

Find the zeros of w^2+7w+12=0

The zeros of w^2+7w+12=0 are w=-3,w=-4

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w^{2} + 7 w + 12 = 0

w = - 3, w = - 4

Solution steps

w^{2} + 7 w + 12 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

w = - 3, w = - 4

6 k - 2 k = 44

3 x^{4} - 48

\frac{3 π}{2} - \frac{π}{4}

8 \cdot 3

x - 2 \geq 5

What are the solutions to the equation w^2+7w+12=0 ?
The solutions to the equation w^2+7w+12=0 are w=-3,w=-4
Find the zeros of w^2+7w+12=0
The zeros of w^2+7w+12=0 are w=-3,w=-4