What are the solutions to the equation x=x^2-4x+3 ?

The solutions to the equation x=x^2-4x+3 are x=(5+sqrt(13))/2 ,x=(5-sqrt(13))/2

Find the zeros of x=x^2-4x+3

The zeros of x=x^2-4x+3 are x=(5+sqrt(13))/2 ,x=(5-sqrt(13))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x = x^{2} - 4 x + 3

x = \frac{5 + 13}{2}, x = \frac{5 - 13}{2}

Decimal Notation

x = 4.30277 \dots, x = 0.69722 \dots

Solution steps

x = x^{2} - 4 x + 3

Switch sides

x^{2} - 4 x + 3 = x

Move

x

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 3 = 13

x_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 13}{2 \cdot 1}

x = \frac{- ( - 5 ) + 13}{2 \cdot 1} : \frac{5 + 13}{2}

x = \frac{- ( - 5 ) - 13}{2 \cdot 1} : \frac{5 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 13}{2}, x = \frac{5 - 13}{2}

3 x - x^{2} = x

a, (a + b + c)^{2} = 144

2 w^{2} + 3 w - 35 = 0

0 = - 12 x^{2} + 4

x, 2 x^{2} + 3 y^{2} + x + 4 y + 5 = 0

What are the solutions to the equation x=x^2-4x+3 ?
The solutions to the equation x=x^2-4x+3 are x=(5+sqrt(13))/2 ,x=(5-sqrt(13))/2
Find the zeros of x=x^2-4x+3
The zeros of x=x^2-4x+3 are x=(5+sqrt(13))/2 ,x=(5-sqrt(13))/2