What are the solutions to the equation d(10d-9)+1=0 ?

The solutions to the equation d(10d-9)+1=0 are d=(9+sqrt(41))/(20),d=(9-sqrt(41))/(20)

Find the zeros of d(10d-9)+1=0

The zeros of d(10d-9)+1=0 are d=(9+sqrt(41))/(20),d=(9-sqrt(41))/(20)

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d (10 d - 9) + 1 = 0

d = \frac{9 + 41}{20}, d = \frac{9 - 41}{20}

Solution steps

d (10 d - 9) + 1 = 0

Expand

d (10 d - 9) + 1 : 10 d^{2} - 9 d + 1

10 d^{2} - 9 d + 1 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 10 \cdot 1}{2 \cdot 10}

(- 9)^{2} - 4 \cdot 10 \cdot 1 = 41

d_{1, 2} = \frac{- ( - 9 ) \pm 41}{2 \cdot 10}

Separate the solutions

d_{1} = \frac{- ( - 9 ) + 41}{2 \cdot 10}, d_{2} = \frac{- ( - 9 ) - 41}{2 \cdot 10}

d = \frac{- ( - 9 ) + 41}{2 \cdot 10} : \frac{9 + 41}{20}

d = \frac{- ( - 9 ) - 41}{2 \cdot 10} : \frac{9 - 41}{20}

The solutions to the quadratic equation are:

d = \frac{9 + 41}{20}, d = \frac{9 - 41}{20}

y^{2} - 4 y = 0

\frac{1000 0 ^{2} \cdot 1.5 \cdot 1 0 ^{- 12}}{2.8}

2 (2 x + 3) = 2 x

\frac{x - 3}{5} \leq - 1

4 (y + 1) + y = 3 (y - 2) + 2

What are the solutions to the equation d(10d-9)+1=0 ?
The solutions to the equation d(10d-9)+1=0 are d=(9+sqrt(41))/(20),d=(9-sqrt(41))/(20)
Find the zeros of d(10d-9)+1=0
The zeros of d(10d-9)+1=0 are d=(9+sqrt(41))/(20),d=(9-sqrt(41))/(20)