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

The solutions to the equation x^2-4x-8=0 are x=2(1+sqrt(3)),x=2(1-sqrt(3))

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

The zeros of x^2-4x-8=0 are x=2(1+sqrt(3)),x=2(1-sqrt(3))

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

x = 2 (1 + 3), x = 2 (1 - 3)

Decimal Notation

x = 5.46410 \dots, x = - 1.46410 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 8) = 43

x_{1, 2} = \frac{- ( - 4 ) \pm 4 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 3}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 4 3}{2 \cdot 1}

x = \frac{- ( - 4 ) + 4 3}{2 \cdot 1} : 2 (1 + 3)

x = \frac{- ( - 4 ) - 4 3}{2 \cdot 1} : 2 (1 - 3)

The solutions to the quadratic equation are:

x = 2 (1 + 3), x = 2 (1 - 3)

x^{2} - 10 x + 25 - y^{2}

\frac{1}{3} (x + 4) - \frac{5}{6} (x - 3) > \frac{1}{2} x + 1

x^{2} - 12 \leq 4 x

\frac{m ^{2} - 7 m - 18}{m ^{2} - 10 m + 9} + \frac{6}{m - 1}

\frac{( x + 6 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{36}

What are the solutions to the equation x^2-4x-8=0 ?
The solutions to the equation x^2-4x-8=0 are x=2(1+sqrt(3)),x=2(1-sqrt(3))
Find the zeros of x^2-4x-8=0
The zeros of x^2-4x-8=0 are x=2(1+sqrt(3)),x=2(1-sqrt(3))