What are the solutions to the equation 4x^2+19x+19=0 ?

The solutions to the equation 4x^2+19x+19=0 are x=(-19+sqrt(57))/8 ,x=(-19-sqrt(57))/8

Find the zeros of 4x^2+19x+19=0

The zeros of 4x^2+19x+19=0 are x=(-19+sqrt(57))/8 ,x=(-19-sqrt(57))/8

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4 x^{2} + 19 x + 19 = 0

x = \frac{- 19 + 57}{8}, x = \frac{- 19 - 57}{8}

Solution steps

4 x^{2} + 19 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 4 \cdot 19}{2 \cdot 4}

1 9^{2} - 4 \cdot 4 \cdot 19 = 57

x_{1, 2} = \frac{- 19 \pm 57}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 19 + 57}{2 \cdot 4}, x_{2} = \frac{- 19 - 57}{2 \cdot 4}

x = \frac{- 19 + 57}{2 \cdot 4} : \frac{- 19 + 57}{8}

x = \frac{- 19 - 57}{2 \cdot 4} : \frac{- 19 - 57}{8}

The solutions to the quadratic equation are:

x = \frac{- 19 + 57}{8}, x = \frac{- 19 - 57}{8}

4 x^{2} + 27 = - 20 x

x^{2} - 2 x + 1

9.08 = 0.001 (0.723 x^{2} + 15.418 x + 607.733)

(6 a + 2) (8 a + 7) = 0

x^{2} - 9 x = 0

What are the solutions to the equation 4x^2+19x+19=0 ?
The solutions to the equation 4x^2+19x+19=0 are x=(-19+sqrt(57))/8 ,x=(-19-sqrt(57))/8
Find the zeros of 4x^2+19x+19=0
The zeros of 4x^2+19x+19=0 are x=(-19+sqrt(57))/8 ,x=(-19-sqrt(57))/8