What are the solutions to the equation 25x^2+18x-2=0 ?

The solutions to the equation 25x^2+18x-2=0 are x=(-9+sqrt(131))/(25),x=-(9+sqrt(131))/(25)

Find the zeros of 25x^2+18x-2=0

The zeros of 25x^2+18x-2=0 are x=(-9+sqrt(131))/(25),x=-(9+sqrt(131))/(25)

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25 x^{2} + 18 x - 2 = 0

x = \frac{- 9 + 131}{25}, x = - \frac{9 + 131}{25}

Decimal Notation

x = 0.09782 \dots, x = - 0.81782 \dots

Solution steps

25 x^{2} + 18 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 25 ( - 2 )}{2 \cdot 25}

1 8^{2} - 4 \cdot 25 (- 2) = 2131

x_{1, 2} = \frac{- 18 \pm 2 131}{2 \cdot 25}

Separate the solutions

x_{1} = \frac{- 18 + 2 131}{2 \cdot 25}, x_{2} = \frac{- 18 - 2 131}{2 \cdot 25}

x = \frac{- 18 + 2 131}{2 \cdot 25} : \frac{- 9 + 131}{25}

x = \frac{- 18 - 2 131}{2 \cdot 25} : - \frac{9 + 131}{25}

The solutions to the quadratic equation are:

x = \frac{- 9 + 131}{25}, x = - \frac{9 + 131}{25}

x^{2} + 15 x - 30 = 0

x^{2} - 3 (2 x - 5) = 10

x (2 - x) = - 2

x, x^{2} - k x + 1681 = 0

(2 x - 1)^{2} = (x + 1)^{2}

What are the solutions to the equation 25x^2+18x-2=0 ?
The solutions to the equation 25x^2+18x-2=0 are x=(-9+sqrt(131))/(25),x=-(9+sqrt(131))/(25)
Find the zeros of 25x^2+18x-2=0
The zeros of 25x^2+18x-2=0 are x=(-9+sqrt(131))/(25),x=-(9+sqrt(131))/(25)