What are the solutions to the equation (x)=x^2+4x+5 ?

The solutions to the equation (x)=x^2+4x+5 are x=-3/2+i(sqrt(11))/2 ,x=-3/2-i(sqrt(11))/2

Find the zeros of (x)=x^2+4x+5

The zeros of (x)=x^2+4x+5 are x=-3/2+i(sqrt(11))/2 ,x=-3/2-i(sqrt(11))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x) = x^{2} + 4 x + 5

x = - \frac{3}{2} + i \frac{11}{2}, x = - \frac{3}{2} - i \frac{11}{2}

Solution steps

(x) = x^{2} + 4 x + 5

Refine

x = x^{2} + 4 x + 5

Switch sides

x^{2} + 4 x + 5 = x

Move

x

to the left side

x^{2} + 3 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

3^{2} - 4 \cdot 1 \cdot 5 : 11 i

x_{1, 2} = \frac{- 3 \pm 11 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 11 i}{2 \cdot 1}, x_{2} = \frac{- 3 - 11 i}{2 \cdot 1}

x = \frac{- 3 + 11 i}{2 \cdot 1} : - \frac{3}{2} + i \frac{11}{2}

x = \frac{- 3 - 11 i}{2 \cdot 1} : - \frac{3}{2} - i \frac{11}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2} + i \frac{11}{2}, x = - \frac{3}{2} - i \frac{11}{2}

(2 x + a)^{2} = b

x^{2} + 8 x = 14

29 - 5 x = x^{2} - 2 x + 1

x^{2} + 8 x = 38

x, 2 x^{2} + 4 x - 7 = 0

What are the solutions to the equation (x)=x^2+4x+5 ?
The solutions to the equation (x)=x^2+4x+5 are x=-3/2+i(sqrt(11))/2 ,x=-3/2-i(sqrt(11))/2
Find the zeros of (x)=x^2+4x+5
The zeros of (x)=x^2+4x+5 are x=-3/2+i(sqrt(11))/2 ,x=-3/2-i(sqrt(11))/2