What are the solutions to the equation -2w^2+12w+19=0 ?

The solutions to the equation -2w^2+12w+19=0 are w=-(-6+sqrt(74))/2 ,w=(6+sqrt(74))/2

Find the zeros of -2w^2+12w+19=0

The zeros of -2w^2+12w+19=0 are w=-(-6+sqrt(74))/2 ,w=(6+sqrt(74))/2

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- 2 w^{2} + 12 w + 19 = 0

w = - \frac{- 6 + 74}{2}, w = \frac{6 + 74}{2}

Decimal

w = - 1.30116 \dots, w = 7.30116 \dots

Solution steps

- 2 w^{2} + 12 w + 19 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 2 ) \cdot 19}{2 ( - 2 )}

1 2^{2} - 4 (- 2) \cdot 19 = 274

w_{1, 2} = \frac{- 12 \pm 2 74}{2 ( - 2 )}

Separate the solutions

w_{1} = \frac{- 12 + 2 74}{2 ( - 2 )}, w_{2} = \frac{- 12 - 2 74}{2 ( - 2 )}

w = \frac{- 12 + 2 74}{2 ( - 2 )} : - \frac{- 6 + 74}{2}

w = \frac{- 12 - 2 74}{2 ( - 2 )} : \frac{6 + 74}{2}

The solutions to the quadratic equation are:

w = - \frac{- 6 + 74}{2}, w = \frac{6 + 74}{2}

y^{2} + x = 1, y

y^{2} + 5 y - 1

x^{2} + 4 x = - 3

- 6 x^{2} + 3 x + 2 = 3

1 = 0.001 x^{2} - 0.612 x + 168

What are the solutions to the equation -2w^2+12w+19=0 ?
The solutions to the equation -2w^2+12w+19=0 are w=-(-6+sqrt(74))/2 ,w=(6+sqrt(74))/2
Find the zeros of -2w^2+12w+19=0
The zeros of -2w^2+12w+19=0 are w=-(-6+sqrt(74))/2 ,w=(6+sqrt(74))/2