What are the solutions to the equation 4x^2-4x-8=0 ?

The solutions to the equation 4x^2-4x-8=0 are x=2,x=-1

Find the zeros of 4x^2-4x-8=0

The zeros of 4x^2-4x-8=0 are x=2,x=-1

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4 x^{2} - 4 x - 8 = 0

x = 2, x = - 1

Solution steps

4 x^{2} - 4 x - 8 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 4 (- 8) = 12

x_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 12}{2 \cdot 4}, x_{2} = \frac{- ( - 4 ) - 12}{2 \cdot 4}

x = \frac{- ( - 4 ) + 12}{2 \cdot 4} : 2

x = \frac{- ( - 4 ) - 12}{2 \cdot 4} : - 1

The solutions to the quadratic equation are:

x = 2, x = - 1

x, f = x^{2} + y^{2} + z^{2}

- 2 x^{2} + 2 x - 3 = 0

- 36 + (x - 12)^{2} = 24

a, a^{2} = b^{2}

x^{2} + 5 x - 12 = 0

What are the solutions to the equation 4x^2-4x-8=0 ?
The solutions to the equation 4x^2-4x-8=0 are x=2,x=-1
Find the zeros of 4x^2-4x-8=0
The zeros of 4x^2-4x-8=0 are x=2,x=-1