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

The solutions to the equation 3d^2+d=-2d are d=0,d=-1

Find the zeros of 3d^2+d=-2d

The zeros of 3d^2+d=-2d are d=0,d=-1

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3 d^{2} + d = - 2 d

d = 0, d = - 1

Solution steps

3 d^{2} + d = - 2 d

Move

2 d

to the left side

3 d^{2} + 3 d = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

3^{2} - 4 \cdot 3 \cdot 0 = 3

d_{1, 2} = \frac{- 3 \pm 3}{2 \cdot 3}

Separate the solutions

d_{1} = \frac{- 3 + 3}{2 \cdot 3}, d_{2} = \frac{- 3 - 3}{2 \cdot 3}

d = \frac{- 3 + 3}{2 \cdot 3} : 0

d = \frac{- 3 - 3}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

d = 0, d = - 1

- 1 < 2 x - 5 < 7

(0.5 a - 0.6 b + 5.5) - (- 0.5 a + 0.4 b) + (1.3 b - 4.5)

x^{2} + 6 x = 14

14 m^{2} - 12 m = 0

710 = 83 + 0.003 p^{2}

What are the solutions to the equation 3d^2+d=-2d ?
The solutions to the equation 3d^2+d=-2d are d=0,d=-1
Find the zeros of 3d^2+d=-2d
The zeros of 3d^2+d=-2d are d=0,d=-1