What are the solutions to the equation 2x^2+4x-22=0 ?

The solutions to the equation 2x^2+4x-22=0 are x=-1+2sqrt(3),x=-1-2sqrt(3)

Find the zeros of 2x^2+4x-22=0

The zeros of 2x^2+4x-22=0 are x=-1+2sqrt(3),x=-1-2sqrt(3)

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2 x^{2} + 4 x - 22 = 0

x = - 1 + 23, x = - 1 - 23

Decimal Notation

x = 2.46410 \dots, x = - 4.46410 \dots

Solution steps

2 x^{2} + 4 x - 22 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 (- 22) = 83

x_{1, 2} = \frac{- 4 \pm 8 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 8 3}{2 \cdot 2}, x_{2} = \frac{- 4 - 8 3}{2 \cdot 2}

x = \frac{- 4 + 8 3}{2 \cdot 2} : - 1 + 23

x = \frac{- 4 - 8 3}{2 \cdot 2} : - 1 - 23

The solutions to the quadratic equation are:

x = - 1 + 23, x = - 1 - 23

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

x, 2 x^{2} + 5 x = 8

y - 3 = y^{2} - 2 y - 13

(3 x + 4) (5 - \frac{x}{2}) = 0

x^{2} - 9 x = 8

What are the solutions to the equation 2x^2+4x-22=0 ?
The solutions to the equation 2x^2+4x-22=0 are x=-1+2sqrt(3),x=-1-2sqrt(3)
Find the zeros of 2x^2+4x-22=0
The zeros of 2x^2+4x-22=0 are x=-1+2sqrt(3),x=-1-2sqrt(3)