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

The solutions to the equation x^2+8x-4=0 are x=2(sqrt(5)-2),x=-2(2+sqrt(5))

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

The zeros of x^2+8x-4=0 are x=2(sqrt(5)-2),x=-2(2+sqrt(5))

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

x = 2 (5 - 2), x = - 2 (2 + 5)

Decimal Notation

x = 0.47213 \dots, x = - 8.47213 \dots

Solution steps

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

Solve with the quadratic formula

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

8^{2} - 4 \cdot 1 \cdot (- 4) = 45

x_{1, 2} = \frac{- 8 \pm 4 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 4 5}{2 \cdot 1}, x_{2} = \frac{- 8 - 4 5}{2 \cdot 1}

x = \frac{- 8 + 4 5}{2 \cdot 1} : 2 (5 - 2)

x = \frac{- 8 - 4 5}{2 \cdot 1} : - 2 (2 + 5)

The solutions to the quadratic equation are:

x = 2 (5 - 2), x = - 2 (2 + 5)

- \frac{1}{2} y^{2} + \frac{9}{8}

3 (25 m^{6}) (\frac{3 n ^{2}}{5})

\frac{x}{2} + \frac{x}{3} + \frac{x}{4} + \frac{x}{6} = 30

l, p_{n} = l + \frac{s i}{f}

- 15 x + 60 \leq 105

What are the solutions to the equation x^2+8x-4=0 ?
The solutions to the equation x^2+8x-4=0 are x=2(sqrt(5)-2),x=-2(2+sqrt(5))
Find the zeros of x^2+8x-4=0
The zeros of x^2+8x-4=0 are x=2(sqrt(5)-2),x=-2(2+sqrt(5))