What are the solutions to the equation 3x^2-19x+3=0 ?

The solutions to the equation 3x^2-19x+3=0 are x=(19+5sqrt(13))/6 ,x=(19-5sqrt(13))/6

Find the zeros of 3x^2-19x+3=0

The zeros of 3x^2-19x+3=0 are x=(19+5sqrt(13))/6 ,x=(19-5sqrt(13))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 x^{2} - 19 x + 3 = 0

x = \frac{19 + 5 13}{6}, x = \frac{19 - 5 13}{6}

Decimal Notation

x = 6.17129 \dots, x = 0.16204 \dots

Solution steps

3 x^{2} - 19 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 19)^{2} - 4 \cdot 3 \cdot 3 = 513

x_{1, 2} = \frac{- ( - 19 ) \pm 5 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 19 ) + 5 13}{2 \cdot 3}, x_{2} = \frac{- ( - 19 ) - 5 13}{2 \cdot 3}

x = \frac{- ( - 19 ) + 5 13}{2 \cdot 3} : \frac{19 + 5 13}{6}

x = \frac{- ( - 19 ) - 5 13}{2 \cdot 3} : \frac{19 - 5 13}{6}

The solutions to the quadratic equation are:

x = \frac{19 + 5 13}{6}, x = \frac{19 - 5 13}{6}

5 u^{2} - 9 uv + 4 v^{2}

\frac{1}{4} - \frac{3}{5}

\frac{2 + 3 i}{- 1 + i}

324 a^{3} b^{7}

5 (15 x + 6) > 3 - 4 x

What are the solutions to the equation 3x^2-19x+3=0 ?
The solutions to the equation 3x^2-19x+3=0 are x=(19+5sqrt(13))/6 ,x=(19-5sqrt(13))/6
Find the zeros of 3x^2-19x+3=0
The zeros of 3x^2-19x+3=0 are x=(19+5sqrt(13))/6 ,x=(19-5sqrt(13))/6