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(sqrt(5)-1)

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(sqrt(5)-1)

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

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

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 ( - 1 ) \cdot 16}{2 ( - 1 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

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

x^{2} = \frac{25}{4}

9 - 15 x^{2} = 13 + 7 x^{2}

16 d^{2} - 81 = 0

0 = - x^{2} - 4

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(sqrt(5)-1)
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(sqrt(5)-1)