What are the solutions to the equation w^2+7w+4=0 ?

The solutions to the equation w^2+7w+4=0 are w=(-7+sqrt(33))/2 ,w=(-7-sqrt(33))/2

Find the zeros of w^2+7w+4=0

The zeros of w^2+7w+4=0 are w=(-7+sqrt(33))/2 ,w=(-7-sqrt(33))/2

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w^{2} + 7 w + 4 = 0

w = \frac{- 7 + 33}{2}, w = \frac{- 7 - 33}{2}

Decimal Notation

w = - 0.62771 \dots, w = - 6.37228 \dots

Solution steps

w^{2} + 7 w + 4 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 1 \cdot 4 = 33

w_{1, 2} = \frac{- 7 \pm 33}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 7 + 33}{2 \cdot 1}, w_{2} = \frac{- 7 - 33}{2 \cdot 1}

w = \frac{- 7 + 33}{2 \cdot 1} : \frac{- 7 + 33}{2}

w = \frac{- 7 - 33}{2 \cdot 1} : \frac{- 7 - 33}{2}

The solutions to the quadratic equation are:

w = \frac{- 7 + 33}{2}, w = \frac{- 7 - 33}{2}

z^{2} + 0.35 z - 1.125 = 0

x^{2} - 7 x - 1 = - 1

x^{2} + 6 x + 8

(x + 4) (2 x - a) = 0

6 y^{2} - 3 y - 4 = y^{2}

What are the solutions to the equation w^2+7w+4=0 ?
The solutions to the equation w^2+7w+4=0 are w=(-7+sqrt(33))/2 ,w=(-7-sqrt(33))/2
Find the zeros of w^2+7w+4=0
The zeros of w^2+7w+4=0 are w=(-7+sqrt(33))/2 ,w=(-7-sqrt(33))/2