What are the solutions to the equation 2w^2+w-28=0 ?

The solutions to the equation 2w^2+w-28=0 are w= 7/2 ,w=-4

Find the zeros of 2w^2+w-28=0

The zeros of 2w^2+w-28=0 are w= 7/2 ,w=-4

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2 w^{2} + w - 28 = 0

w = \frac{7}{2}, w = - 4

Decimal Notation

w = 3.5, w = - 4

Solution steps

2 w^{2} + w - 28 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 28 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 28) = 15

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

Separate the solutions

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

w = \frac{- 1 + 15}{2 \cdot 2} : \frac{7}{2}

w = \frac{- 1 - 15}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

w = \frac{7}{2}, w = - 4

15 x^{3} (- 11 x^{2} y^{2})

2 z - 42 \geq 0

(6 u)^{2}

∣ 3 x + 2 ∣ = ∣ x - 11 ∣

- 83 = 2 u - 15 u - 18

What are the solutions to the equation 2w^2+w-28=0 ?
The solutions to the equation 2w^2+w-28=0 are w= 7/2 ,w=-4
Find the zeros of 2w^2+w-28=0
The zeros of 2w^2+w-28=0 are w= 7/2 ,w=-4