What are the solutions to the equation w^2+50=-15w ?

The solutions to the equation w^2+50=-15w are w=-5,w=-10

Find the zeros of w^2+50=-15w

The zeros of w^2+50=-15w are w=-5,w=-10

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w^{2} + 50 = - 15 w

w = - 5, w = - 10

Solution steps

w^{2} + 50 = - 15 w

Move

15 w

to the left side

w^{2} + 50 + 15 w = 0

Write in the standard form

a x^{2} + b x + c = 0

w^{2} + 15 w + 50 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 50}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 50 = 5

w_{1, 2} = \frac{- 15 \pm 5}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 15 + 5}{2 \cdot 1}, w_{2} = \frac{- 15 - 5}{2 \cdot 1}

w = \frac{- 15 + 5}{2 \cdot 1} : - 5

w = \frac{- 15 - 5}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

w = - 5, w = - 10

4 x^{2} + 7 x - 7 = 0

3 x^{2} - 5 x + 1

(6 x - 3) (2 x + 4) = 3

4 x^{2} - 2 x + 10 = 0

x^{2} + 9 x = 27

What are the solutions to the equation w^2+50=-15w ?
The solutions to the equation w^2+50=-15w are w=-5,w=-10
Find the zeros of w^2+50=-15w
The zeros of w^2+50=-15w are w=-5,w=-10