What are the solutions to the equation x^2-4x-1=3 ?

The solutions to the equation x^2-4x-1=3 are x=2(1+sqrt(2)),x=2(1-sqrt(2))

Find the zeros of x^2-4x-1=3

The zeros of x^2-4x-1=3 are x=2(1+sqrt(2)),x=2(1-sqrt(2))

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

x = 2 (1 + 2), x = 2 (1 - 2)

Decimal Notation

x = 4.82842 \dots, x = - 0.82842 \dots

Solution steps

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

Move

3

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 4) = 42

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 (1 + 2), x = 2 (1 - 2)

- x^{2} - 2 x + 24 = 0

x^{2} - 4 + 6 = 0

x^{2} = - 5 x - 3

x^{2} = lo g_{2} (48) - lo g_{2} (3)

x^{2} = 4 x - 13

What are the solutions to the equation x^2-4x-1=3 ?
The solutions to the equation x^2-4x-1=3 are x=2(1+sqrt(2)),x=2(1-sqrt(2))
Find the zeros of x^2-4x-1=3
The zeros of x^2-4x-1=3 are x=2(1+sqrt(2)),x=2(1-sqrt(2))