What are the solutions to the equation d^2-4d+5=d ?

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

Find the zeros of d^2-4d+5=d

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

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d^{2} - 4 d + 5 = d

d = \frac{5 + 5}{2}, d = \frac{5 - 5}{2}

Decimal Notation

d = 3.61803 \dots, d = 1.38196 \dots

Solution steps

d^{2} - 4 d + 5 = d

Move

d

to the left side

d^{2} - 5 d + 5 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 1 \cdot 5 = 5

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

d = \frac{5 + 5}{2}, d = \frac{5 - 5}{2}

(8 y^{3}) (- 3 x^{2} y^{2}) (\frac{3}{8} x y^{4})

x = 22 - 33 e^{- l n (\frac{10}{33}) (4)}

- 3 (- w - 3) = 3 w - 26

2 x^{2} + 3 x + 5 \leq 0

5 x - 2 > 3 x + 6

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