What are the solutions to the equation 6d^2-3d-4=d^2 ?

The solutions to the equation 6d^2-3d-4=d^2 are d=(3+sqrt(89))/(10),d=(3-sqrt(89))/(10)

Find the zeros of 6d^2-3d-4=d^2

The zeros of 6d^2-3d-4=d^2 are d=(3+sqrt(89))/(10),d=(3-sqrt(89))/(10)

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6 d^{2} - 3 d - 4 = d^{2}

d = \frac{3 + 89}{10}, d = \frac{3 - 89}{10}

Decimal Notation

d = 1.24339 \dots, d = - 0.64339 \dots

Solution steps

6 d^{2} - 3 d - 4 = d^{2}

Move

d^{2}

to the left side

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 5 (- 4) = 89

d_{1, 2} = \frac{- ( - 3 ) \pm 89}{2 \cdot 5}

Separate the solutions

d_{1} = \frac{- ( - 3 ) + 89}{2 \cdot 5}, d_{2} = \frac{- ( - 3 ) - 89}{2 \cdot 5}

d = \frac{- ( - 3 ) + 89}{2 \cdot 5} : \frac{3 + 89}{10}

d = \frac{- ( - 3 ) - 89}{2 \cdot 5} : \frac{3 - 89}{10}

The solutions to the quadratic equation are:

d = \frac{3 + 89}{10}, d = \frac{3 - 89}{10}

3 = 20 \cdot \frac{2}{2} \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^{2}

(x - 1)^{2} - 2 = 0

- 3 x^{2} + 48 x = 0

1 = \frac{x ^{2}}{2} + 4

- 8 x = 2 x^{2}

What are the solutions to the equation 6d^2-3d-4=d^2 ?
The solutions to the equation 6d^2-3d-4=d^2 are d=(3+sqrt(89))/(10),d=(3-sqrt(89))/(10)
Find the zeros of 6d^2-3d-4=d^2
The zeros of 6d^2-3d-4=d^2 are d=(3+sqrt(89))/(10),d=(3-sqrt(89))/(10)