What are the solutions to the equation 3+5d=4d^2 ?

The solutions to the equation 3+5d=4d^2 are d=(5+sqrt(73))/8 ,d=(5-sqrt(73))/8

Find the zeros of 3+5d=4d^2

The zeros of 3+5d=4d^2 are d=(5+sqrt(73))/8 ,d=(5-sqrt(73))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 + 5 d = 4 d^{2}

d = \frac{5 + 73}{8}, d = \frac{5 - 73}{8}

Decimal Notation

d = 1.69300 \dots, d = - 0.44300 \dots

Solution steps

3 + 5 d = 4 d^{2}

Switch sides

4 d^{2} = 3 + 5 d

Move

5 d

to the left side

4 d^{2} - 5 d = 3

Move

3

to the left side

4 d^{2} - 5 d - 3 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

(- 5)^{2} - 4 \cdot 4 (- 3) = 73

d_{1, 2} = \frac{- ( - 5 ) \pm 73}{2 \cdot 4}

Separate the solutions

d_{1} = \frac{- ( - 5 ) + 73}{2 \cdot 4}, d_{2} = \frac{- ( - 5 ) - 73}{2 \cdot 4}

d = \frac{- ( - 5 ) + 73}{2 \cdot 4} : \frac{5 + 73}{8}

d = \frac{- ( - 5 ) - 73}{2 \cdot 4} : \frac{5 - 73}{8}

The solutions to the quadratic equation are:

d = \frac{5 + 73}{8}, d = \frac{5 - 73}{8}

(2 r - 5)^{2} = 0

21 k^{2} + 4 k = - 3 + 6 k + 9 k^{2}

3 a^{2} - 27 = 0

x^{2} - 8 x = 9 x - 12

4 x - 10 (- x - 1) = 6 (x - 2) 6 (x - 2)

What are the solutions to the equation 3+5d=4d^2 ?
The solutions to the equation 3+5d=4d^2 are d=(5+sqrt(73))/8 ,d=(5-sqrt(73))/8
Find the zeros of 3+5d=4d^2
The zeros of 3+5d=4d^2 are d=(5+sqrt(73))/8 ,d=(5-sqrt(73))/8