What are the solutions to the equation 2x^2+12x-18=0 ?

The solutions to the equation 2x^2+12x-18=0 are x=3(sqrt(2)-1),x=-3(1+sqrt(2))

Find the zeros of 2x^2+12x-18=0

The zeros of 2x^2+12x-18=0 are x=3(sqrt(2)-1),x=-3(1+sqrt(2))

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2 x^{2} + 12 x - 18 = 0

x = 3 (2 - 1), x = - 3 (1 + 2)

Solution steps

2 x^{2} + 12 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 ( - 18 )}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 (- 18) = 122

x_{1, 2} = \frac{- 12 \pm 12 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 12 2}{2 \cdot 2}, x_{2} = \frac{- 12 - 12 2}{2 \cdot 2}

x = \frac{- 12 + 12 2}{2 \cdot 2} : 3 (2 - 1)

x = \frac{- 12 - 12 2}{2 \cdot 2} : - 3 (1 + 2)

The solutions to the quadratic equation are:

x = 3 (2 - 1), x = - 3 (1 + 2)

3^{- x} > 6^{- x}

36 y^{2} - 12 y + 1 = 0

z^{3} + 9 z^{2} + 20 z

lo g_{10} (∣ x - 1 ∣) = 3

2^{2 x} - 5 (2^{x}) + 4 < 0

What are the solutions to the equation 2x^2+12x-18=0 ?
The solutions to the equation 2x^2+12x-18=0 are x=3(sqrt(2)-1),x=-3(1+sqrt(2))
Find the zeros of 2x^2+12x-18=0
The zeros of 2x^2+12x-18=0 are x=3(sqrt(2)-1),x=-3(1+sqrt(2))