What are the solutions to the equation 2x^2+20x-8=0 ?

The solutions to the equation 2x^2+20x-8=0 are x=-5+sqrt(29),x=-5-sqrt(29)

Find the zeros of 2x^2+20x-8=0

The zeros of 2x^2+20x-8=0 are x=-5+sqrt(29),x=-5-sqrt(29)

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2 x^{2} + 20 x - 8 = 0

x = - 5 + 29, x = - 5 - 29

Decimal Notation

x = 0.38516 \dots, x = - 10.38516 \dots

Solution steps

2 x^{2} + 20 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 2 ( - 8 )}{2 \cdot 2}

2 0^{2} - 4 \cdot 2 (- 8) = 429

x_{1, 2} = \frac{- 20 \pm 4 29}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 20 + 4 29}{2 \cdot 2}, x_{2} = \frac{- 20 - 4 29}{2 \cdot 2}

x = \frac{- 20 + 4 29}{2 \cdot 2} : - 5 + 29

x = \frac{- 20 - 4 29}{2 \cdot 2} : - 5 - 29

The solutions to the quadratic equation are:

x = - 5 + 29, x = - 5 - 29

321.0069 x^{2} - 5130.1778 x + 14682.6031 = 0

x (5 x - 2) = 24

x^{2} - 18 x + 4 = 0

- x^{2} - 12 x + 9 = 0

x, z = x^{2} y^{2} - x y

What are the solutions to the equation 2x^2+20x-8=0 ?
The solutions to the equation 2x^2+20x-8=0 are x=-5+sqrt(29),x=-5-sqrt(29)
Find the zeros of 2x^2+20x-8=0
The zeros of 2x^2+20x-8=0 are x=-5+sqrt(29),x=-5-sqrt(29)