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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Decimal Notation

x = 2.47213 \dots, x = - 6.47213 \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 (5 - 1)

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

The solutions to the quadratic equation are:

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

(x + 4) (x - 4) = 0

32 x^{2} - 4 x - 15

- 9 d - 4 (- 3 d - 3) \geq 6 d + 6 + 6 d

∣ x - 9 ∣ = - 5

8 - 2 n = 2

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