What are the solutions to the equation d^2-2d-4=0 ?

The solutions to the equation d^2-2d-4=0 are d=1+sqrt(5),d=1-sqrt(5)

Find the zeros of d^2-2d-4=0

The zeros of d^2-2d-4=0 are d=1+sqrt(5),d=1-sqrt(5)

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

d = 1 + 5, d = 1 - 5

Decimal Notation

d = 3.23606 \dots, d = - 1.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

d = 1 + 5, d = 1 - 5

x, x^{2} - k x + 1 = 0

3 x^{2} - 8 x - 10 = 4

3 x^{2} + 3 x - 9 = 4 x - 7

3 b^{2} + 9 b - 20 = 2 b

2 x - 3 = - 5 x^{2}

What are the solutions to the equation d^2-2d-4=0 ?
The solutions to the equation d^2-2d-4=0 are d=1+sqrt(5),d=1-sqrt(5)
Find the zeros of d^2-2d-4=0
The zeros of d^2-2d-4=0 are d=1+sqrt(5),d=1-sqrt(5)