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

The solutions to the equation x^2-2x-8=4 are x=1+sqrt(13),x=1-sqrt(13)

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

The zeros of x^2-2x-8=4 are x=1+sqrt(13),x=1-sqrt(13)

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

x = 1 + 13, x = 1 - 13

Decimal Notation

x = 4.60555 \dots, x = - 2.60555 \dots

Solution steps

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

Move

4

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 12) = 213

x_{1, 2} = \frac{- ( - 2 ) \pm 2 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 13}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 13}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 13}{2 \cdot 1} : 1 + 13

x = \frac{- ( - 2 ) - 2 13}{2 \cdot 1} : 1 - 13

The solutions to the quadratic equation are:

x = 1 + 13, x = 1 - 13

196 x^{2} - 1 = 0

x, x^{2} - 5 x - 4 = 0

3 x^{2} - 4 x + 1 = 0

n (n + 1) = 2

- 8 z = 12 + z^{2}

What are the solutions to the equation x^2-2x-8=4 ?
The solutions to the equation x^2-2x-8=4 are x=1+sqrt(13),x=1-sqrt(13)
Find the zeros of x^2-2x-8=4
The zeros of x^2-2x-8=4 are x=1+sqrt(13),x=1-sqrt(13)