What are the solutions to the equation (2x-7)^2+16=0 ?

The solutions to the equation (2x-7)^2+16=0 are x= 7/2+2i,x= 7/2-2i

Find the zeros of (2x-7)^2+16=0

The zeros of (2x-7)^2+16=0 are x= 7/2+2i,x= 7/2-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2 x - 7)^{2} + 16 = 0

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

Solution steps

(2 x - 7)^{2} + 16 = 0

Move

16

to the right side

(2 x - 7)^{2} = - 16

For

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

the solutions are

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

Solve

2 x - 7 = - 16 : x = \frac{7}{2} + 2 i

Solve

2 x - 7 = - - 16 : x = \frac{7}{2} - 2 i

The solutions to the quadratic equation are:

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

10 x^{2} + 4 x + 3920 = 0

2 m^{2} + 4 = 0

x, 4 x^{2} + k x - 6 = 3

5 x^{2} + 20 x + 265 = 0

(x + 2) x - 3 (x - 2) = (x + 1) 2 - (x - 3) x

What are the solutions to the equation (2x-7)^2+16=0 ?
The solutions to the equation (2x-7)^2+16=0 are x= 7/2+2i,x= 7/2-2i
Find the zeros of (2x-7)^2+16=0
The zeros of (2x-7)^2+16=0 are x= 7/2+2i,x= 7/2-2i