What are the solutions to the equation 3x+1=2x^2-3x+1 ?

The solutions to the equation 3x+1=2x^2-3x+1 are x=3,x=0

Find the zeros of 3x+1=2x^2-3x+1

The zeros of 3x+1=2x^2-3x+1 are x=3,x=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 x + 1 = 2 x^{2} - 3 x + 1

x = 3, x = 0

Solution steps

3 x + 1 = 2 x^{2} - 3 x + 1

Switch sides

2 x^{2} - 3 x + 1 = 3 x + 1

Move

1

to the left side

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

Move

3 x

to the left side

2 x^{2} - 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 6)^{2} - 4 \cdot 2 \cdot 0 = 6

x_{1, 2} = \frac{- ( - 6 ) \pm 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 6}{2 \cdot 2}, x_{2} = \frac{- ( - 6 ) - 6}{2 \cdot 2}

x = \frac{- ( - 6 ) + 6}{2 \cdot 2} : 3

x = \frac{- ( - 6 ) - 6}{2 \cdot 2} : 0

The solutions to the quadratic equation are:

x = 3, x = 0

4 t = 2 t^{2}

\frac{- 9 x ^{2} - 240 x}{80} + 60 = 0

a, (a^{3} b^{5}) (a^{7} b^{5}) = a^{10} b^{10}

2 r^{2} + r - 2 = 0

y, x^{2} + 2 y^{2} = 6

What are the solutions to the equation 3x+1=2x^2-3x+1 ?
The solutions to the equation 3x+1=2x^2-3x+1 are x=3,x=0
Find the zeros of 3x+1=2x^2-3x+1
The zeros of 3x+1=2x^2-3x+1 are x=3,x=0