What are the solutions to the equation (y+1)^2-20(y+1)=-64 ?

The solutions to the equation (y+1)^2-20(y+1)=-64 are y=15,y=3

Find the zeros of (y+1)^2-20(y+1)=-64

The zeros of (y+1)^2-20(y+1)=-64 are y=15,y=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y + 1)^{2} - 20 (y + 1) = - 64

y = 15, y = 3

Solution steps

(y + 1)^{2} - 20 (y + 1) = - 64

Expand

(y + 1)^{2} - 20 (y + 1) : y^{2} - 18 y - 19

y^{2} - 18 y - 19 = - 64

Move

64

to the left side

y^{2} - 18 y + 45 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 45}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 45 = 12

y_{1, 2} = \frac{- ( - 18 ) \pm 12}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 18 ) + 12}{2 \cdot 1}, y_{2} = \frac{- ( - 18 ) - 12}{2 \cdot 1}

y = \frac{- ( - 18 ) + 12}{2 \cdot 1} : 15

y = \frac{- ( - 18 ) - 12}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

y = 15, y = 3

97 \leq - 8 x + 1

3 (x - 3) - 16 = 2 (x - 8) + 8

2 x - 5 = 13

x x^{8}

(a - 2 b)^{2} - (a + 2 b)^{2}

What are the solutions to the equation (y+1)^2-20(y+1)=-64 ?
The solutions to the equation (y+1)^2-20(y+1)=-64 are y=15,y=3
Find the zeros of (y+1)^2-20(y+1)=-64
The zeros of (y+1)^2-20(y+1)=-64 are y=15,y=3