What are the solutions to the equation 0.5x(x+1)=-1 ?

The solutions to the equation 0.5x(x+1)=-1 are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2

Find the zeros of 0.5x(x+1)=-1

The zeros of 0.5x(x+1)=-1 are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0.5 x (x + 1) = - 1

x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

Solution steps

0.5 x (x + 1) = - 1

Multiply both sides by

10

5 x (x + 1) = - 10

Expand

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

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

Move

10

to the left side

5 x^{2} + 5 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 5 \cdot 10}{2 \cdot 5}

Simplify

5^{2} - 4 \cdot 5 \cdot 10 : 57 i

x_{1, 2} = \frac{- 5 \pm 5 7 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 5 + 5 7 i}{2 \cdot 5}, x_{2} = \frac{- 5 - 5 7 i}{2 \cdot 5}

x = \frac{- 5 + 5 7 i}{2 \cdot 5} : - \frac{1}{2} + i \frac{7}{2}

x = \frac{- 5 - 5 7 i}{2 \cdot 5} : - \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

d^{0.4} x (x - 1) - 3 (x^{2} - 5) - x^{2} = (x - 3) - (x + 4)

x, f (x + h) = x^{2} + h

t, x = v_{0} \cdot t + \frac{1}{2} \cdot a \cdot t^{2}

- 332.65 = (0.0004) x^{2} - 0.7281 x

x, 3 x^{2} - 18 y^{2} + 2 z^{2} + 3 y^{2} z^{2} - 18 x = 27

What are the solutions to the equation 0.5x(x+1)=-1 ?
The solutions to the equation 0.5x(x+1)=-1 are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2
Find the zeros of 0.5x(x+1)=-1
The zeros of 0.5x(x+1)=-1 are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2