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

The solutions to the equation 4d^2-d-3=-d^2 are d=(1+sqrt(61))/(10),d=(1-sqrt(61))/(10)

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

The zeros of 4d^2-d-3=-d^2 are d=(1+sqrt(61))/(10),d=(1-sqrt(61))/(10)

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

d = \frac{1 + 61}{10}, d = \frac{1 - 61}{10}

Decimal

d = 0.88102 \dots, d = - 0.68102 \dots

Solution steps

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

Move

d^{2}

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 5 (- 3) = 61

d_{1, 2} = \frac{- ( - 1 ) \pm 61}{2 \cdot 5}

Separate the solutions

d_{1} = \frac{- ( - 1 ) + 61}{2 \cdot 5}, d_{2} = \frac{- ( - 1 ) - 61}{2 \cdot 5}

d = \frac{- ( - 1 ) + 61}{2 \cdot 5} : \frac{1 + 61}{10}

d = \frac{- ( - 1 ) - 61}{2 \cdot 5} : \frac{1 - 61}{10}

The solutions to the quadratic equation are:

d = \frac{1 + 61}{10}, d = \frac{1 - 61}{10}

2 x^{2} - x - 10 = 0

(x) = x^{2} - 6 x + 8

(k + 2) x^{2} + 2 k x + k + 1 = 0

x^{2} - 12 x + 33 = 0

(a - 3)^{2} = 12

What are the solutions to the equation 4d^2-d-3=-d^2 ?
The solutions to the equation 4d^2-d-3=-d^2 are d=(1+sqrt(61))/(10),d=(1-sqrt(61))/(10)
Find the zeros of 4d^2-d-3=-d^2
The zeros of 4d^2-d-3=-d^2 are d=(1+sqrt(61))/(10),d=(1-sqrt(61))/(10)