What are the solutions to the equation 5d^2+6d+1=4d^2 ?

The solutions to the equation 5d^2+6d+1=4d^2 are d=-3+2sqrt(2),d=-3-2sqrt(2)

Find the zeros of 5d^2+6d+1=4d^2

The zeros of 5d^2+6d+1=4d^2 are d=-3+2sqrt(2),d=-3-2sqrt(2)

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

d = - 3 + 22, d = - 3 - 22

Decimal Notation

d = - 0.17157 \dots, d = - 5.82842 \dots

Solution steps

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

Move

4 d^{2}

to the left side

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

Solve with the quadratic formula

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

6^{2} - 4 \cdot 1 \cdot 1 = 42

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

Separate the solutions

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

d = \frac{- 6 + 4 2}{2 \cdot 1} : - 3 + 22

d = \frac{- 6 - 4 2}{2 \cdot 1} : - 3 - 22

The solutions to the quadratic equation are:

d = - 3 + 22, d = - 3 - 22

3 x^{2} - 4 = 140

3 x^{2} - 10 = 38

x, x^{2} + y^{2} - 2 x + 2 y + 2 = 0

x, y = (x - h)^{2} + k

- 0.0002037 x^{2} + 0.0170086 x - 0.06426 = 0

What are the solutions to the equation 5d^2+6d+1=4d^2 ?
The solutions to the equation 5d^2+6d+1=4d^2 are d=-3+2sqrt(2),d=-3-2sqrt(2)
Find the zeros of 5d^2+6d+1=4d^2
The zeros of 5d^2+6d+1=4d^2 are d=-3+2sqrt(2),d=-3-2sqrt(2)