What are the solutions to the equation e=d^2*10+d^1 ?

The solutions to the equation e=d^2*10+d^1 are d=(-1+sqrt(1+40e))/(20),d=(-1-sqrt(1+40e))/(20)

Find the zeros of e=d^2*10+d^1

The zeros of e=d^2*10+d^1 are d=(-1+sqrt(1+40e))/(20),d=(-1-sqrt(1+40e))/(20)

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e = d^{2} \cdot 10 + d^{1}

d = \frac{- 1 + 1 + 40 e}{20}, d = \frac{- 1 - 1 + 40 e}{20}

Decimal Notation

d = 0.47376 \dots, d = - 0.57376 \dots

Solution steps

e = d^{2} \cdot 10 + d^{1}

Refine

e = 10 d^{2} + d

Switch sides

10 d^{2} + d = e

Move

e

to the left side

10 d^{2} + d - e = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 10 (- e) = 1 + 40 e

d_{1, 2} = \frac{- 1 \pm 1 + 40 e}{2 \cdot 10}

Separate the solutions

d_{1} = \frac{- 1 + 1 + 40 e}{2 \cdot 10}, d_{2} = \frac{- 1 - 1 + 40 e}{2 \cdot 10}

d = \frac{- 1 + 1 + 40 e}{2 \cdot 10} : \frac{- 1 + 1 + 40 e}{20}

d = \frac{- 1 - 1 + 40 e}{2 \cdot 10} : \frac{- 1 - 1 + 40 e}{20}

The solutions to the quadratic equation are:

d = \frac{- 1 + 1 + 40 e}{20}, d = \frac{- 1 - 1 + 40 e}{20}

y, x^{3} + y^{2} - 27 x - 10 y + 30 = 0

(x - 4)^{2} - 1 = - 0.75

2 x - 7 = x^{2} + 1

n^{2} - 52 + 12 = 0

3 x^{2} - (m + 2) x + 4 m - 2 = 0

What are the solutions to the equation e=d^2*10+d^1 ?
The solutions to the equation e=d^2*10+d^1 are d=(-1+sqrt(1+40e))/(20),d=(-1-sqrt(1+40e))/(20)
Find the zeros of e=d^2*10+d^1
The zeros of e=d^2*10+d^1 are d=(-1+sqrt(1+40e))/(20),d=(-1-sqrt(1+40e))/(20)