What are the solutions to the equation d^2+10d+28=3 ?

The solutions to the equation d^2+10d+28=3 are d=-5

Find the zeros of d^2+10d+28=3

The zeros of d^2+10d+28=3 are d=-5

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d^{2} + 10 d + 28 = 3

d = - 5

Solution steps

d^{2} + 10 d + 28 = 3

Move

3

to the left side

d^{2} + 10 d + 25 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 25}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot 25 = 0

d_{1, 2} = \frac{- 10 \pm 0}{2 \cdot 1}

d = \frac{- 10}{2 \cdot 1}

\frac{- 10}{2 \cdot 1} = - 5

d = - 5

The solution to the quadratic equation is:

d = - 5

x^{2} + 61 = 1 - 17 x

9 x^{2} + 19 = 24 x

- 3 x^{2} + 5 x - 11 = 0

n^{2} + 11 n + 3 = - 4 n^{2}

20 - x^{2} = 10

What are the solutions to the equation d^2+10d+28=3 ?
The solutions to the equation d^2+10d+28=3 are d=-5
Find the zeros of d^2+10d+28=3
The zeros of d^2+10d+28=3 are d=-5