What are the solutions to the equation 3(3+3)=(y+1)^2 ?

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

Find the zeros of 3(3+3)=(y+1)^2

The zeros of 3(3+3)=(y+1)^2 are y=3sqrt(2)-1,y=-3sqrt(2)-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 (3 + 3) = (y + 1)^{2}

y = 32 - 1, y = - 32 - 1

Decimal Notation

y = 3.24264 \dots, y = - 5.24264 \dots

Solution steps

3 (3 + 3) = (y + 1)^{2}

Switch sides

(y + 1)^{2} = 3 (3 + 3)

Simplify

3 (3 + 3) : 18

For

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

the solutions are

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

Solve

y + 1 = 18 : y = 32 - 1

Solve

y + 1 = - 18 : y = - 32 - 1

The solutions to the quadratic equation are:

y = 32 - 1, y = - 32 - 1

70 (60 s^{2} + 40 s - 0.5) = 0

10 8^{2}

5 x^{2} - 50 x + 120

2 x - 1 \geq 7

- \frac{1}{2} (350) (0.2)^{2}

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