What are the solutions to the equation 12x^2-13x+4=13x^2-5x+1 ?

The solutions to the equation 12x^2-13x+4=13x^2-5x+1 are x=-4-sqrt(19),x=sqrt(19)-4

Find the zeros of 12x^2-13x+4=13x^2-5x+1

The zeros of 12x^2-13x+4=13x^2-5x+1 are x=-4-sqrt(19),x=sqrt(19)-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12 x^{2} - 13 x + 4 = 13 x^{2} - 5 x + 1

x = - 4 - 19, x = 19 - 4

Decimal Notation

x = - 8.35889 \dots, x = 0.35889 \dots

Solution steps

12 x^{2} - 13 x + 4 = 13 x^{2} - 5 x + 1

Move

1

to the left side

12 x^{2} - 13 x + 3 = 13 x^{2} - 5 x

Move

5 x

to the left side

12 x^{2} - 8 x + 3 = 13 x^{2}

Move

13 x^{2}

to the left side

- x^{2} - 8 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

(- 8)^{2} - 4 (- 1) \cdot 3 = 219

x_{1, 2} = \frac{- ( - 8 ) \pm 2 19}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 19}{2 ( - 1 )}, x_{2} = \frac{- ( - 8 ) - 2 19}{2 ( - 1 )}

x = \frac{- ( - 8 ) + 2 19}{2 ( - 1 )} : - 4 - 19

x = \frac{- ( - 8 ) - 2 19}{2 ( - 1 )} : 19 - 4

The solutions to the quadratic equation are:

x = - 4 - 19, x = 19 - 4

x^{2} + 5 x - 310 = 0

a^{2} - 8 a - 12 = 0

y, (x + y)^{3} = x^{3} + y^{3}

x^{2} + 28 x + c

r, α r^{2} + r + α r + 1 = 0

What are the solutions to the equation 12x^2-13x+4=13x^2-5x+1 ?
The solutions to the equation 12x^2-13x+4=13x^2-5x+1 are x=-4-sqrt(19),x=sqrt(19)-4
Find the zeros of 12x^2-13x+4=13x^2-5x+1
The zeros of 12x^2-13x+4=13x^2-5x+1 are x=-4-sqrt(19),x=sqrt(19)-4