What are the solutions to the equation x^2+12+20x=0 ?

The solutions to the equation x^2+12+20x=0 are x=2(sqrt(22)-5),x=-2(5+sqrt(22))

Find the zeros of x^2+12+20x=0

The zeros of x^2+12+20x=0 are x=2(sqrt(22)-5),x=-2(5+sqrt(22))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + 12 + 20 x = 0

x = 2 (22 - 5), x = - 2 (5 + 22)

Decimal Notation

x = - 0.61916 \dots, x = - 19.38083 \dots

Solution steps

x^{2} + 12 + 20 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + 20 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot 12 = 422

x_{1, 2} = \frac{- 20 \pm 4 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 4 22}{2 \cdot 1}, x_{2} = \frac{- 20 - 4 22}{2 \cdot 1}

x = \frac{- 20 + 4 22}{2 \cdot 1} : 2 (22 - 5)

x = \frac{- 20 - 4 22}{2 \cdot 1} : - 2 (5 + 22)

The solutions to the quadratic equation are:

x = 2 (22 - 5), x = - 2 (5 + 22)

9 x^{2} - 6 x + 38 = 0

s^{2} - 4 s + 13

2 x^{2} - 6 x = 6 x^{2} - 8 x

x, 2 z = x^{2} + y^{2}

4 x^{2} + 4 x + 1 = 121

What are the solutions to the equation x^2+12+20x=0 ?
The solutions to the equation x^2+12+20x=0 are x=2(sqrt(22)-5),x=-2(5+sqrt(22))
Find the zeros of x^2+12+20x=0
The zeros of x^2+12+20x=0 are x=2(sqrt(22)-5),x=-2(5+sqrt(22))