What are the solutions to the equation -3x^2-4x+12=0 ?

The solutions to the equation -3x^2-4x+12=0 are x=-(2(1+sqrt(10)))/3 ,x=(2(sqrt(10)-1))/3

Find the zeros of -3x^2-4x+12=0

The zeros of -3x^2-4x+12=0 are x=-(2(1+sqrt(10)))/3 ,x=(2(sqrt(10)-1))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 3 x^{2} - 4 x + 12 = 0

x = - \frac{2 ( 1 + 10 )}{3}, x = \frac{2 ( 10 - 1 )}{3}

Decimal Notation

x = - 2.77485 \dots, x = 1.44151 \dots

Solution steps

- 3 x^{2} - 4 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 3 ) \cdot 12}{2 ( - 3 )}

(- 4)^{2} - 4 (- 3) \cdot 12 = 410

x_{1, 2} = \frac{- ( - 4 ) \pm 4 10}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 10}{2 ( - 3 )}, x_{2} = \frac{- ( - 4 ) - 4 10}{2 ( - 3 )}

x = \frac{- ( - 4 ) + 4 10}{2 ( - 3 )} : - \frac{2 ( 1 + 10 )}{3}

x = \frac{- ( - 4 ) - 4 10}{2 ( - 3 )} : \frac{2 ( 10 - 1 )}{3}

The solutions to the quadratic equation are:

x = - \frac{2 ( 1 + 10 )}{3}, x = \frac{2 ( 10 - 1 )}{3}

y^{2} - 196 = 0

3 r^{2} - 16 r - 2 = 0

2 x^{2} - 21 x + 40 = 0

x^{2} - 5.1 x + 6.3 = 0

\frac{x ^{2}}{3} = 4 x - x^{2}

What are the solutions to the equation -3x^2-4x+12=0 ?
The solutions to the equation -3x^2-4x+12=0 are x=-(2(1+sqrt(10)))/3 ,x=(2(sqrt(10)-1))/3
Find the zeros of -3x^2-4x+12=0
The zeros of -3x^2-4x+12=0 are x=-(2(1+sqrt(10)))/3 ,x=(2(sqrt(10)-1))/3