What are the solutions to the equation d^2+56=-18d ?

The solutions to the equation d^2+56=-18d are d=-4,d=-14

Find the zeros of d^2+56=-18d

The zeros of d^2+56=-18d are d=-4,d=-14

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d^{2} + 56 = - 18 d

d = - 4, d = - 14

Solution steps

d^{2} + 56 = - 18 d

Move

18 d

to the left side

d^{2} + 56 + 18 d = 0

Write in the standard form

a x^{2} + b x + c = 0

d^{2} + 18 d + 56 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 1 \cdot 56}{2 \cdot 1}

1 8^{2} - 4 \cdot 1 \cdot 56 = 10

d_{1, 2} = \frac{- 18 \pm 10}{2 \cdot 1}

Separate the solutions

d_{1} = \frac{- 18 + 10}{2 \cdot 1}, d_{2} = \frac{- 18 - 10}{2 \cdot 1}

d = \frac{- 18 + 10}{2 \cdot 1} : - 4

d = \frac{- 18 - 10}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

d = - 4, d = - 14

3 t^{2} - 12 = 0

2 x - 5 x (2 - x) = 3 x - 7

23.8 = 3.5 b

s^{2} = 20

- 2 x^{2} + 7 x + 5 = 0

What are the solutions to the equation d^2+56=-18d ?
The solutions to the equation d^2+56=-18d are d=-4,d=-14
Find the zeros of d^2+56=-18d
The zeros of d^2+56=-18d are d=-4,d=-14