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

The solutions to the equation 2x^2-2x-4=0 are x=2,x=-1

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

The zeros of 2x^2-2x-4=0 are x=2,x=-1

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

x = 2, x = - 1

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 4) = 6

x_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 2}

x = \frac{- ( - 2 ) + 6}{2 \cdot 2} : 2

x = \frac{- ( - 2 ) - 6}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = 2, x = - 1

81 x^{2} - 49

5 x + 4 (x - 1) = - 2 x + \frac{1}{4}

\frac{x ^{2} + 2 x + 1}{x ^{2} - 1}

\frac{- 7 n ^{2}}{- 14 n ^{2} - 35 n}

1 0^{2} + b^{2} = 2 6^{2}

What are the solutions to the equation 2x^2-2x-4=0 ?
The solutions to the equation 2x^2-2x-4=0 are x=2,x=-1
Find the zeros of 2x^2-2x-4=0
The zeros of 2x^2-2x-4=0 are x=2,x=-1