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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Multiply both sides by

2

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

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

9 x^{2} - 8 = 0

x, x^{2} y^{2} + x y = 1

25 - r^{2} = 4 r

x^{2} + 4 = 2 x^{2} - 12

3 x^{2} - 3 = 45

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