What are the solutions to the equation (x+1)^2+(x+2)^2=-10 ?

The solutions to the equation (x+1)^2+(x+2)^2=-10 are x=-3/2+i(sqrt(21))/2 ,x=-3/2-i(sqrt(21))/2

Find the zeros of (x+1)^2+(x+2)^2=-10

The zeros of (x+1)^2+(x+2)^2=-10 are x=-3/2+i(sqrt(21))/2 ,x=-3/2-i(sqrt(21))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 1)^{2} + (x + 2)^{2} = - 10

x = - \frac{3}{2} + i \frac{21}{2}, x = - \frac{3}{2} - i \frac{21}{2}

Solution steps

(x + 1)^{2} + (x + 2)^{2} = - 10

Expand

(x + 1)^{2} + (x + 2)^{2} : 2 x^{2} + 6 x + 5

2 x^{2} + 6 x + 5 = - 10

Move

10

to the left side

2 x^{2} + 6 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 \cdot 15}{2 \cdot 2}

Simplify

6^{2} - 4 \cdot 2 \cdot 15 : 221 i

x_{1, 2} = \frac{- 6 \pm 2 21 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 2 21 i}{2 \cdot 2}, x_{2} = \frac{- 6 - 2 21 i}{2 \cdot 2}

x = \frac{- 6 + 2 21 i}{2 \cdot 2} : - \frac{3}{2} + i \frac{21}{2}

x = \frac{- 6 - 2 21 i}{2 \cdot 2} : - \frac{3}{2} - i \frac{21}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2} + i \frac{21}{2}, x = - \frac{3}{2} - i \frac{21}{2}

2 x - 5 \geq 8 x + 10

x^{2} - 6 x + 9 < 16

\frac{2}{3} x = 6

h^{2} - 2 h - 2 = 0

m^{2} - 4

What are the solutions to the equation (x+1)^2+(x+2)^2=-10 ?
The solutions to the equation (x+1)^2+(x+2)^2=-10 are x=-3/2+i(sqrt(21))/2 ,x=-3/2-i(sqrt(21))/2
Find the zeros of (x+1)^2+(x+2)^2=-10
The zeros of (x+1)^2+(x+2)^2=-10 are x=-3/2+i(sqrt(21))/2 ,x=-3/2-i(sqrt(21))/2