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

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

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

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

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x^{2} - 4 x = 1

x = 2 + 5, x = 2 - 5

Decimal Notation

x = 4.23606 \dots, x = - 0.23606 \dots

Solution steps

x^{2} - 4 x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- ( - 4 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 5}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 5}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 5}{2 \cdot 1} : 2 + 5

x = \frac{- ( - 4 ) - 2 5}{2 \cdot 1} : 2 - 5

The solutions to the quadratic equation are:

x = 2 + 5, x = 2 - 5

4 x \leq 124

\frac{1}{( \frac{3}{2} )}

x (7 - 6 x) = (1 + 3 x) (1 + 2 x)

x - 2 4 - x + 1 \geq 0

(a - 1)^{2} - (b - 2)^{2}

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