What are the solutions to the equation (x+1)^2-4x=1 ?

The solutions to the equation (x+1)^2-4x=1 are x=2,x=0

Find the zeros of (x+1)^2-4x=1

The zeros of (x+1)^2-4x=1 are x=2,x=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 1)^{2} - 4 x = 1

x = 2, x = 0

Solution steps

(x + 1)^{2} - 4 x = 1

Expand

(x + 1)^{2} - 4 x : x^{2} - 2 x + 1

x^{2} - 2 x + 1 = 1

Move

1

to the left side

x^{2} - 2 x = 0

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2}{2 \cdot 1} : 2

x = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 2, x = 0

124 = - 16 t^{2} + 100 t + 10

2 x^{2} + 5 x + 2 = 0

x, y = - 48.141 x^{2} + 39.476 x

12 x^{2} - 120 x + 225 = 0

r, V = π \cdot r^{2} \cdot h

What are the solutions to the equation (x+1)^2-4x=1 ?
The solutions to the equation (x+1)^2-4x=1 are x=2,x=0
Find the zeros of (x+1)^2-4x=1
The zeros of (x+1)^2-4x=1 are x=2,x=0