What are the solutions to the equation 2x^2+4x-9=0 ?

The solutions to the equation 2x^2+4x-9=0 are x=(-2+sqrt(22))/2 ,x=-(2+sqrt(22))/2

Find the zeros of 2x^2+4x-9=0

The zeros of 2x^2+4x-9=0 are x=(-2+sqrt(22))/2 ,x=-(2+sqrt(22))/2

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2 x^{2} + 4 x - 9 = 0

x = \frac{- 2 + 22}{2}, x = - \frac{2 + 22}{2}

Decimal Notation

x = 1.34520 \dots, x = - 3.34520 \dots

Solution steps

2 x^{2} + 4 x - 9 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 (- 9) = 222

x_{1, 2} = \frac{- 4 \pm 2 22}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 2 22}{2 \cdot 2}, x_{2} = \frac{- 4 - 2 22}{2 \cdot 2}

x = \frac{- 4 + 2 22}{2 \cdot 2} : \frac{- 2 + 22}{2}

x = \frac{- 4 - 2 22}{2 \cdot 2} : - \frac{2 + 22}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 22}{2}, x = - \frac{2 + 22}{2}

5 x - 1 (8 - x) = 10

∣ 5 x - 4 ∣ \geq 6

x^{2} + (0.000447 x) + (- 0.0000129183) = 0

0 > (4 x + 2)^{- 1}

x^{4} - x^{2} + 4 x^{3} - 4 x - 45 x^{2} + 45

What are the solutions to the equation 2x^2+4x-9=0 ?
The solutions to the equation 2x^2+4x-9=0 are x=(-2+sqrt(22))/2 ,x=-(2+sqrt(22))/2
Find the zeros of 2x^2+4x-9=0
The zeros of 2x^2+4x-9=0 are x=(-2+sqrt(22))/2 ,x=-(2+sqrt(22))/2