What are the solutions to the equation 3=(x-1)^2+3(x+2) ?

The solutions to the equation 3=(x-1)^2+3(x+2) are x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2

Find the zeros of 3=(x-1)^2+3(x+2)

The zeros of 3=(x-1)^2+3(x+2) are x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 = (x - 1)^{2} + 3 (x + 2)

x = - \frac{1}{2} + i \frac{15}{2}, x = - \frac{1}{2} - i \frac{15}{2}

Solution steps

3 = (x - 1)^{2} + 3 (x + 2)

Expand

(x - 1)^{2} + 3 (x + 2) : x^{2} + x + 7

3 = x^{2} + x + 7

Switch sides

x^{2} + x + 7 = 3

Move

3

to the left side

x^{2} + x + 4 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 4 : 15 i

x_{1, 2} = \frac{- 1 \pm 15 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 15 i}{2 \cdot 1}, x_{2} = \frac{- 1 - 15 i}{2 \cdot 1}

x = \frac{- 1 + 15 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{15}{2}

x = \frac{- 1 - 15 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{15}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{15}{2}, x = - \frac{1}{2} - i \frac{15}{2}

(7 + 2 x) (3 + 2 x) = 56

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

6 x^{2} - 6 = 5 x

5 x^{2} - 250 x + 3000 = 0

x, 2 x^{2} - 8 x + b y + 10 = 0

What are the solutions to the equation 3=(x-1)^2+3(x+2) ?
The solutions to the equation 3=(x-1)^2+3(x+2) are x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2
Find the zeros of 3=(x-1)^2+3(x+2)
The zeros of 3=(x-1)^2+3(x+2) are x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2