What are the solutions to the equation 2^2+b^2=4^2 ?

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

Find the zeros of 2^2+b^2=4^2

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2^{2} + b^{2} = 4^{2}

b = 23, b = - 23

Decimal Notation

b = 3.46410 \dots, b = - 3.46410 \dots

Solution steps

2^{2} + b^{2} = 4^{2}

Move

2^{2}

to the right side

b^{2} = 12

For

x^{2} = f (a)

the solutions are

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

b = 12, b = - 12

12 = 23

- 12 = - 23

b = 23, b = - 23

2 x^{2} + x - 4 = - 6 x - 7

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

8 x = 3 x^{2} + 5

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

y^{2} - y - 24 = 0

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