What are the solutions to the equation (D+5)3D=-2/4 (6-20D)+3 ?

The solutions to the equation (D+5)3D=-2/4 (6-20D)+3 are D=0,D=-5/3

Find the zeros of (D+5)3D=-2/4 (6-20D)+3

The zeros of (D+5)3D=-2/4 (6-20D)+3 are D=0,D=-5/3

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(D + 5) 3 D = - \frac{2}{4} (6 - 20 D) + 3

D = 0, D = - \frac{5}{3}

Decimal Notation

D = 0, D = - 1.66666 \dots

Solution steps

(D + 5) \cdot 3 D = - \frac{2}{4} (6 - 20 D) + 3

Multiply both sides by

4

12 D (D + 5) = - 2 (6 - 20 D) + 12

Expand

12 D (D + 5) : 12 D^{2} + 60 D

Expand

- 2 (6 - 20 D) + 12 : 40 D

12 D^{2} + 60 D = 40 D

Move

40 D

to the left side

12 D^{2} + 20 D = 0

Solve with the quadratic formula

D_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 12 \cdot 0}{2 \cdot 12}

2 0^{2} - 4 \cdot 12 \cdot 0 = 20

D_{1, 2} = \frac{- 20 \pm 20}{2 \cdot 12}

Separate the solutions

D_{1} = \frac{- 20 + 20}{2 \cdot 12}, D_{2} = \frac{- 20 - 20}{2 \cdot 12}

D = \frac{- 20 + 20}{2 \cdot 12} : 0

D = \frac{- 20 - 20}{2 \cdot 12} : - \frac{5}{3}

The solutions to the quadratic equation are:

D = 0, D = - \frac{5}{3}

2 m^{2} + m = 0

x^{2} = 2 x + 13

x^{2} = 2 x + 25

x^{2} + 9 x + 7 = 0

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

What are the solutions to the equation (D+5)3D=-2/4 (6-20D)+3 ?
The solutions to the equation (D+5)3D=-2/4 (6-20D)+3 are D=0,D=-5/3
Find the zeros of (D+5)3D=-2/4 (6-20D)+3
The zeros of (D+5)3D=-2/4 (6-20D)+3 are D=0,D=-5/3