What are the solutions to the equation r^2=(-2)^2+(2sqrt(3))^2 ?

The solutions to the equation r^2=(-2)^2+(2sqrt(3))^2 are r=4,r=-4

Find the zeros of r^2=(-2)^2+(2sqrt(3))^2

The zeros of r^2=(-2)^2+(2sqrt(3))^2 are r=4,r=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{2} = (- 2)^{2} + (23)^{2}

r = 4, r = - 4

Solution steps

r^{2} = (- 2)^{2} + (23)^{2}

Simplify

(- 2)^{2} + (23)^{2} : 16

For

x^{2} = f (a)

the solutions are

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

r = 16, r = - 16

16 = 4

- 16 = - 4

r = 4, r = - 4

\frac{x ^{2}}{3} + 2 x = x

x^{2} + 20 x - 45 = (x - 4)^{2} - (x - 4)^{2}

9 m^{2} - 6 m + 1 = 0

w^{2} + 375 - 376 = 0

2 t^{2} - 9 t + 4 = 0

What are the solutions to the equation r^2=(-2)^2+(2sqrt(3))^2 ?
The solutions to the equation r^2=(-2)^2+(2sqrt(3))^2 are r=4,r=-4
Find the zeros of r^2=(-2)^2+(2sqrt(3))^2
The zeros of r^2=(-2)^2+(2sqrt(3))^2 are r=4,r=-4