What are the solutions to the equation x^2+8x-4d=0 ?

The solutions to the equation x^2+8x-4d=0 are x=2(sqrt(d+4)-2),x=-2(sqrt(d+4)+2)

Find the zeros of x^2+8x-4d=0

The zeros of x^2+8x-4d=0 are x=2(sqrt(d+4)-2),x=-2(sqrt(d+4)+2)

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x^{2} + 8 x - 4 d = 0

x = 2 (d + 4 - 2), x = - 2 (d + 4 + 2)

Solution steps

x^{2} + 8 x - 4 d = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 4 d )}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot (- 4 d) : 4 4 + d

x_{1, 2} = \frac{- 8 \pm 4 4 + d}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 4 4 + d}{2 \cdot 1}, x_{2} = \frac{- 8 - 4 4 + d}{2 \cdot 1}

x = \frac{- 8 + 4 4 + d}{2 \cdot 1} : 2 (d + 4 - 2)

x = \frac{- 8 - 4 4 + d}{2 \cdot 1} : - 2 (d + 4 + 2)

The solutions to the quadratic equation are:

x = 2 (d + 4 - 2), x = - 2 (d + 4 + 2)

2 x^{2} - 16 x = - 22

6 - 6 x + x^{2} = 0

25 k^{2} + 36 k + 9 = 6 k

a, a^{2} - b^{2} = 0

x^{2} - 2 x = 15

What are the solutions to the equation x^2+8x-4d=0 ?
The solutions to the equation x^2+8x-4d=0 are x=2(sqrt(d+4)-2),x=-2(sqrt(d+4)+2)
Find the zeros of x^2+8x-4d=0
The zeros of x^2+8x-4d=0 are x=2(sqrt(d+4)-2),x=-2(sqrt(d+4)+2)