What are the solutions to the equation 4x^2+32x=-16 ?

The solutions to the equation 4x^2+32x=-16 are x=2(sqrt(3)-2),x=-2(2+sqrt(3))

Find the zeros of 4x^2+32x=-16

The zeros of 4x^2+32x=-16 are x=2(sqrt(3)-2),x=-2(2+sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 x^{2} + 32 x = - 16

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

Decimal Notation

x = - 0.53589 \dots, x = - 7.46410 \dots

Solution steps

4 x^{2} + 32 x = - 16

Move

16

to the left side

4 x^{2} + 32 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

3 2^{2} - 4 \cdot 4 \cdot 16 = 163

x_{1, 2} = \frac{- 32 \pm 16 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 32 + 16 3}{2 \cdot 4}, x_{2} = \frac{- 32 - 16 3}{2 \cdot 4}

x = \frac{- 32 + 16 3}{2 \cdot 4} : 2 (3 - 2)

x = \frac{- 32 - 16 3}{2 \cdot 4} : - 2 (2 + 3)

The solutions to the quadratic equation are:

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

2 x^{2} = - 12 x - 18

- 4 x^{2} + 88 x = 0

2 x^{2} - 2 = 16

x, f = x^{2}

- 3 x^{2} + 9 x - 8 = 0

What are the solutions to the equation 4x^2+32x=-16 ?
The solutions to the equation 4x^2+32x=-16 are x=2(sqrt(3)-2),x=-2(2+sqrt(3))
Find the zeros of 4x^2+32x=-16
The zeros of 4x^2+32x=-16 are x=2(sqrt(3)-2),x=-2(2+sqrt(3))