What are the solutions to the equation (2y-3)^2=5 ?

The solutions to the equation (2y-3)^2=5 are y=(sqrt(5)+3)/2 ,y=(-sqrt(5)+3)/2

Find the zeros of (2y-3)^2=5

The zeros of (2y-3)^2=5 are y=(sqrt(5)+3)/2 ,y=(-sqrt(5)+3)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2 y - 3)^{2} = 5

y = \frac{5 + 3}{2}, y = \frac{- 5 + 3}{2}

Decimal Notation

y = 2.61803 \dots, y = 0.38196 \dots

Solution steps

(2 y - 3)^{2} = 5

For

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

the solutions are

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

Solve

2 y - 3 = 5 : y = \frac{5 + 3}{2}

Solve

2 y - 3 = - 5 : y = \frac{- 5 + 3}{2}

The solutions to the quadratic equation are:

y = \frac{5 + 3}{2}, y = \frac{- 5 + 3}{2}

3 x^{2} - 14 - 5 = 0

x^{2} - 3 x + 3 = 0

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

3 n^{2} = 7 n - 1

7 = 3 x^{2}

What are the solutions to the equation (2y-3)^2=5 ?
The solutions to the equation (2y-3)^2=5 are y=(sqrt(5)+3)/2 ,y=(-sqrt(5)+3)/2
Find the zeros of (2y-3)^2=5
The zeros of (2y-3)^2=5 are y=(sqrt(5)+3)/2 ,y=(-sqrt(5)+3)/2