What are the solutions to the equation (y+5)(y-5)=4 ?

The solutions to the equation (y+5)(y-5)=4 are y=sqrt(29),y=-sqrt(29)

Find the zeros of (y+5)(y-5)=4

The zeros of (y+5)(y-5)=4 are y=sqrt(29),y=-sqrt(29)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y + 5) (y - 5) = 4

y = 29, y = - 29

Decimal Notation

y = 5.38516 \dots, y = - 5.38516 \dots

Solution steps

(y + 5) (y - 5) = 4

Expand

(y + 5) (y - 5) : y^{2} - 25

y^{2} - 25 = 4

Move

4

to the left side

y^{2} - 29 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 29 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 29) = 229

y_{1, 2} = \frac{- 0 \pm 2 29}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 0 + 2 29}{2 \cdot 1}, y_{2} = \frac{- 0 - 2 29}{2 \cdot 1}

y = \frac{- 0 + 2 29}{2 \cdot 1} : 29

y = \frac{- 0 - 2 29}{2 \cdot 1} : - 29

The solutions to the quadratic equation are:

y = 29, y = - 29

x, x^{2} y + y = 1

8 y^{2} - 15 = 4 y^{2} + 1

3 x (2 x - 1) = 6

2 (3 x + 1)^{2} = 0

- 2 x^{2} + 2 x + 40 = 0

What are the solutions to the equation (y+5)(y-5)=4 ?
The solutions to the equation (y+5)(y-5)=4 are y=sqrt(29),y=-sqrt(29)
Find the zeros of (y+5)(y-5)=4
The zeros of (y+5)(y-5)=4 are y=sqrt(29),y=-sqrt(29)