What are the solutions to the equation 18x^2+19x-7=0 ?

The solutions to the equation 18x^2+19x-7=0 are x=(-19+sqrt(865))/(36),x=(-19-sqrt(865))/(36)

Find the zeros of 18x^2+19x-7=0

The zeros of 18x^2+19x-7=0 are x=(-19+sqrt(865))/(36),x=(-19-sqrt(865))/(36)

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18 x^{2} + 19 x - 7 = 0

x = \frac{- 19 + 865}{36}, x = \frac{- 19 - 865}{36}

Decimal Notation

x = 0.28919 \dots, x = - 1.34474 \dots

Solution steps

18 x^{2} + 19 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 18 ( - 7 )}{2 \cdot 18}

1 9^{2} - 4 \cdot 18 (- 7) = 865

x_{1, 2} = \frac{- 19 \pm 865}{2 \cdot 18}

Separate the solutions

x_{1} = \frac{- 19 + 865}{2 \cdot 18}, x_{2} = \frac{- 19 - 865}{2 \cdot 18}

x = \frac{- 19 + 865}{2 \cdot 18} : \frac{- 19 + 865}{36}

x = \frac{- 19 - 865}{2 \cdot 18} : \frac{- 19 - 865}{36}

The solutions to the quadratic equation are:

x = \frac{- 19 + 865}{36}, x = \frac{- 19 - 865}{36}

3 x (x - 3) = 6 x^{2} - 11 x

x \cdot x = 5 x

x \cdot x = 36

- 2 x + x^{2} = 15

x, (x - 7)^{2} + (x - 7) - 6 = 0

What are the solutions to the equation 18x^2+19x-7=0 ?
The solutions to the equation 18x^2+19x-7=0 are x=(-19+sqrt(865))/(36),x=(-19-sqrt(865))/(36)
Find the zeros of 18x^2+19x-7=0
The zeros of 18x^2+19x-7=0 are x=(-19+sqrt(865))/(36),x=(-19-sqrt(865))/(36)