What are the solutions to the equation d^2+d-19=4d ?

The solutions to the equation d^2+d-19=4d are d=(3+sqrt(85))/2 ,d=(3-sqrt(85))/2

Find the zeros of d^2+d-19=4d

The zeros of d^2+d-19=4d are d=(3+sqrt(85))/2 ,d=(3-sqrt(85))/2

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d^{2} + d - 19 = 4 d

d = \frac{3 + 85}{2}, d = \frac{3 - 85}{2}

Decimal Notation

d = 6.10977 \dots, d = - 3.10977 \dots

Solution steps

d^{2} + d - 19 = 4 d

Move

4 d

to the left side

d^{2} - 3 d - 19 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 19 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 19) = 85

d_{1, 2} = \frac{- ( - 3 ) \pm 85}{2 \cdot 1}

Separate the solutions

d_{1} = \frac{- ( - 3 ) + 85}{2 \cdot 1}, d_{2} = \frac{- ( - 3 ) - 85}{2 \cdot 1}

d = \frac{- ( - 3 ) + 85}{2 \cdot 1} : \frac{3 + 85}{2}

d = \frac{- ( - 3 ) - 85}{2 \cdot 1} : \frac{3 - 85}{2}

The solutions to the quadratic equation are:

d = \frac{3 + 85}{2}, d = \frac{3 - 85}{2}

m^{2} + 11 m = - 30

x^{(2)} + 5 x + 6 = 0

x, f (a + h) = - 5 x^{2} + 2 x - 1

x, f (x)^{2} = x + 2

x^{1} - x^{2} = 4

What are the solutions to the equation d^2+d-19=4d ?
The solutions to the equation d^2+d-19=4d are d=(3+sqrt(85))/2 ,d=(3-sqrt(85))/2
Find the zeros of d^2+d-19=4d
The zeros of d^2+d-19=4d are d=(3+sqrt(85))/2 ,d=(3-sqrt(85))/2