What are the solutions to the equation 1/3 d^2+1/2 d=-1/12 ?

The solutions to the equation 1/3 d^2+1/2 d=-1/12 are d=(-3+sqrt(5))/4 ,d=-(3+sqrt(5))/4

Find the zeros of 1/3 d^2+1/2 d=-1/12

The zeros of 1/3 d^2+1/2 d=-1/12 are d=(-3+sqrt(5))/4 ,d=-(3+sqrt(5))/4

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\frac{1}{3} d^{2} + \frac{1}{2} d = - \frac{1}{12}

d = \frac{- 3 + 5}{4}, d = - \frac{3 + 5}{4}

Solution steps

\frac{1}{3} d^{2} + \frac{1}{2} d = - \frac{1}{12}

Find Least Common Multiplier of

3, 2, 12 : 12

Multiply by LCM=

12

\frac{1}{3} d^{2} \cdot 12 + \frac{1}{2} d \cdot 12 = - \frac{1}{12} \cdot 12

Simplify

4 d^{2} + 6 d = - 1

Move

1

to the left side

4 d^{2} + 6 d + 1 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

6^{2} - 4 \cdot 4 \cdot 1 = 25

d_{1, 2} = \frac{- 6 \pm 2 5}{2 \cdot 4}

Separate the solutions

d_{1} = \frac{- 6 + 2 5}{2 \cdot 4}, d_{2} = \frac{- 6 - 2 5}{2 \cdot 4}

d = \frac{- 6 + 2 5}{2 \cdot 4} : \frac{- 3 + 5}{4}

d = \frac{- 6 - 2 5}{2 \cdot 4} : - \frac{3 + 5}{4}

The solutions to the quadratic equation are:

d = \frac{- 3 + 5}{4}, d = - \frac{3 + 5}{4}

y^{2} - 14 y = 0

(16 x^{8} y^{- 12})^{\frac{1}{2}}

8 x^{2} + 40 x + 50

4^{2} + 9 (4) + 2

∣ x ∣ + 4 = 2

What are the solutions to the equation 1/3 d^2+1/2 d=-1/12 ?
The solutions to the equation 1/3 d^2+1/2 d=-1/12 are d=(-3+sqrt(5))/4 ,d=-(3+sqrt(5))/4
Find the zeros of 1/3 d^2+1/2 d=-1/12
The zeros of 1/3 d^2+1/2 d=-1/12 are d=(-3+sqrt(5))/4 ,d=-(3+sqrt(5))/4