What are the solutions to the equation x^2-4x-16=0 ?

The solutions to the equation x^2-4x-16=0 are x=2(1+sqrt(5)),x=2(1-sqrt(5))

Find the zeros of x^2-4x-16=0

The zeros of x^2-4x-16=0 are x=2(1+sqrt(5)),x=2(1-sqrt(5))

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x^{2} - 4 x - 16 = 0

x = 2 (1 + 5), x = 2 (1 - 5)

Decimal Notation

x = 6.47213 \dots, x = - 2.47213 \dots

Solution steps

x^{2} - 4 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 16 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 16) = 45

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 (1 + 5), x = 2 (1 - 5)

\frac{x - 1}{x - 5} \leq 2

(\frac{125}{x ^{6}})^{\frac{1}{3}}

\frac{6 x + 1}{x - 3} \leq 0

16 - y^{2}

h^{2} + 18 h - 19

What are the solutions to the equation x^2-4x-16=0 ?
The solutions to the equation x^2-4x-16=0 are x=2(1+sqrt(5)),x=2(1-sqrt(5))
Find the zeros of x^2-4x-16=0
The zeros of x^2-4x-16=0 are x=2(1+sqrt(5)),x=2(1-sqrt(5))