What are the solutions to the equation 4B=(t+2)^2 ?

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

Find the zeros of 4B=(t+2)^2

The zeros of 4B=(t+2)^2 are t=2sqrt(B)-2,t=-2sqrt(B)-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 B = (t + 2)^{2}

t = 2 B - 2, t = - 2 B - 2

Solution steps

4 B = (t + 2)^{2}

Switch sides

(t + 2)^{2} = 4 B

For

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

the solutions are

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

Solve

t + 2 = 4 B : t = 2 B - 2

Solve

t + 2 = - 4 B : t = - 2 B - 2

The solutions to the quadratic equation are:

t = 2 B - 2, t = - 2 B - 2

6 a^{2} - 2 = 0

x, g = a (x - h)^{2} + 3

\frac{3}{2} x^{2} = x + 4

- 108 + x^{2} = 0

- 16 t^{2} + 1250 = - 16 (6)^{2} + 1250

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