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

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

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

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

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

x = 4, x = - 2

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 8) = 6

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 6}{2 \cdot 1} : 4

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

The solutions to the quadratic equation are:

x = 4, x = - 2

(\frac{1}{2})^{x + 3}

(x + 9)^{2} (x + 4) > 0

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

c^{4} - 1

(\frac{16}{3})^{2}

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