What are the solutions to the equation 120w^2+1800w-4080=0 ?

The solutions to the equation 120w^2+1800w-4080=0 are w=2,w=-17

Find the zeros of 120w^2+1800w-4080=0

The zeros of 120w^2+1800w-4080=0 are w=2,w=-17

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120 w^{2} + 1800 w - 4080 = 0

w = 2, w = - 17

Solution steps

120 w^{2} + 1800 w - 4080 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 1800 \pm 180 0 ^{2} - 4 \cdot 120 ( - 4080 )}{2 \cdot 120}

180 0^{2} - 4 \cdot 120 (- 4080) = 2280

w_{1, 2} = \frac{- 1800 \pm 2280}{2 \cdot 120}

Separate the solutions

w_{1} = \frac{- 1800 + 2280}{2 \cdot 120}, w_{2} = \frac{- 1800 - 2280}{2 \cdot 120}

w = \frac{- 1800 + 2280}{2 \cdot 120} : 2

w = \frac{- 1800 - 2280}{2 \cdot 120} : - 17

The solutions to the quadratic equation are:

w = 2, w = - 17

t, - 0.8 t^{2} + 32 t = 0

- 4.9 x^{2} - 3 x + 30

5 x^{2} + 30 x + 45 = 0

y^{2} - 5 y = 2 y

- 3 x^{2} - 12 x - 5 = 0

What are the solutions to the equation 120w^2+1800w-4080=0 ?
The solutions to the equation 120w^2+1800w-4080=0 are w=2,w=-17
Find the zeros of 120w^2+1800w-4080=0
The zeros of 120w^2+1800w-4080=0 are w=2,w=-17