What are the solutions to the equation 0=x^2+10x+29 ?

The solutions to the equation 0=x^2+10x+29 are x=-5+2i,x=-5-2i

Find the zeros of 0=x^2+10x+29

The zeros of 0=x^2+10x+29 are x=-5+2i,x=-5-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = x^{2} + 10 x + 29

x = - 5 + 2 i, x = - 5 - 2 i

Solution steps

0 = x^{2} + 10 x + 29

Switch sides

x^{2} + 10 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

1 0^{2} - 4 \cdot 1 \cdot 29 : 4 i

x_{1, 2} = \frac{- 10 \pm 4 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 4 i}{2 \cdot 1}, x_{2} = \frac{- 10 - 4 i}{2 \cdot 1}

x = \frac{- 10 + 4 i}{2 \cdot 1} : - 5 + 2 i

x = \frac{- 10 - 4 i}{2 \cdot 1} : - 5 - 2 i

The solutions to the quadratic equation are:

x = - 5 + 2 i, x = - 5 - 2 i

5 (4 n - 4) = - 60

x^{2} + 6 < 2 x

218 - 212 + 218

x = x^{2} - 2

3 x + 1 < 7

What are the solutions to the equation 0=x^2+10x+29 ?
The solutions to the equation 0=x^2+10x+29 are x=-5+2i,x=-5-2i
Find the zeros of 0=x^2+10x+29
The zeros of 0=x^2+10x+29 are x=-5+2i,x=-5-2i