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

The solutions to the equation 2w^2+w-7=0 are w=(-1+sqrt(57))/4 ,w=(-1-sqrt(57))/4

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

The zeros of 2w^2+w-7=0 are w=(-1+sqrt(57))/4 ,w=(-1-sqrt(57))/4

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

w = \frac{- 1 + 57}{4}, w = \frac{- 1 - 57}{4}

Decimal Notation

w = 1.63745 \dots, w = - 2.13745 \dots

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 7) = 57

w_{1, 2} = \frac{- 1 \pm 57}{2 \cdot 2}

Separate the solutions

w_{1} = \frac{- 1 + 57}{2 \cdot 2}, w_{2} = \frac{- 1 - 57}{2 \cdot 2}

w = \frac{- 1 + 57}{2 \cdot 2} : \frac{- 1 + 57}{4}

w = \frac{- 1 - 57}{2 \cdot 2} : \frac{- 1 - 57}{4}

The solutions to the quadratic equation are:

w = \frac{- 1 + 57}{4}, w = \frac{- 1 - 57}{4}

3 x^{2} - 3 x - 10 = 0

4 x^{2} + 5 x + 5 = 0

(x - 3)^{2} + 5 = - 11

4 v^{2} - 9 v + 4 = 0

3 r^{2} + 75 = 0

What are the solutions to the equation 2w^2+w-7=0 ?
The solutions to the equation 2w^2+w-7=0 are w=(-1+sqrt(57))/4 ,w=(-1-sqrt(57))/4
Find the zeros of 2w^2+w-7=0
The zeros of 2w^2+w-7=0 are w=(-1+sqrt(57))/4 ,w=(-1-sqrt(57))/4