What are the solutions to the equation 2x^2+20x-22=0 ?

The solutions to the equation 2x^2+20x-22=0 are x=1,x=-11

Find the zeros of 2x^2+20x-22=0

The zeros of 2x^2+20x-22=0 are x=1,x=-11

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2 x^{2} + 20 x - 22 = 0

x = 1, x = - 11

Solution steps

2 x^{2} + 20 x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 2 ( - 22 )}{2 \cdot 2}

2 0^{2} - 4 \cdot 2 (- 22) = 24

x_{1, 2} = \frac{- 20 \pm 24}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 20 + 24}{2 \cdot 2}, x_{2} = \frac{- 20 - 24}{2 \cdot 2}

x = \frac{- 20 + 24}{2 \cdot 2} : 1

x = \frac{- 20 - 24}{2 \cdot 2} : - 11

The solutions to the quadratic equation are:

x = 1, x = - 11

2 x^{2} - 28 x + 102

x, x^{2} + 2 x - 24 = 0

15 x = x^{2} - 16

0 = - \frac{1}{2} (x - 2)^{2} + 8

- 0.000003 x^{2} + 0.003 x + 6 = 0

What are the solutions to the equation 2x^2+20x-22=0 ?
The solutions to the equation 2x^2+20x-22=0 are x=1,x=-11
Find the zeros of 2x^2+20x-22=0
The zeros of 2x^2+20x-22=0 are x=1,x=-11