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

The solutions to the equation x^2-2x=4 are x=1+sqrt(5),x=1-sqrt(5)

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

The zeros of x^2-2x=4 are x=1+sqrt(5),x=1-sqrt(5)

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x^{2} - 2 x = 4

x = 1 + 5, x = 1 - 5

Decimal Notation

x = 3.23606 \dots, x = - 1.23606 \dots

Solution steps

x^{2} - 2 x = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 4) = 25

x_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 1}

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

x = \frac{- ( - 2 ) - 2 5}{2 \cdot 1} : 1 - 5

The solutions to the quadratic equation are:

x = 1 + 5, x = 1 - 5

5 - x = 3

8 = \frac{k}{5}

3 (2 x + 5) - 2 (4 + 4 x) = 7

x^{2} + π x + 2 = 0

What are the solutions to the equation x^2-2x=4 ?
The solutions to the equation x^2-2x=4 are x=1+sqrt(5),x=1-sqrt(5)
Find the zeros of x^2-2x=4
The zeros of x^2-2x=4 are x=1+sqrt(5),x=1-sqrt(5)