What are the solutions to the equation -3d^2-4d+5=0 ?

The solutions to the equation -3d^2-4d+5=0 are d=-(2+sqrt(19))/3 ,d=(sqrt(19)-2)/3

Find the zeros of -3d^2-4d+5=0

The zeros of -3d^2-4d+5=0 are d=-(2+sqrt(19))/3 ,d=(sqrt(19)-2)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 3 d^{2} - 4 d + 5 = 0

d = - \frac{2 + 19}{3}, d = \frac{19 - 2}{3}

Decimal Notation

d = - 2.11963 \dots, d = 0.78629 \dots

Solution steps

- 3 d^{2} - 4 d + 5 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 3) \cdot 5 = 219

d_{1, 2} = \frac{- ( - 4 ) \pm 2 19}{2 ( - 3 )}

Separate the solutions

d_{1} = \frac{- ( - 4 ) + 2 19}{2 ( - 3 )}, d_{2} = \frac{- ( - 4 ) - 2 19}{2 ( - 3 )}

d = \frac{- ( - 4 ) + 2 19}{2 ( - 3 )} : - \frac{2 + 19}{3}

d = \frac{- ( - 4 ) - 2 19}{2 ( - 3 )} : \frac{19 - 2}{3}

The solutions to the quadratic equation are:

d = - \frac{2 + 19}{3}, d = \frac{19 - 2}{3}

25 r^{2} + 7 = 107

6 x^{2} + 5 x + 8 = 0

a, (a + b)^{3} = a^{3} + b^{3}

x^{2} = 3 x - 4

x, (x - 5)^{2} - (y + 3)^{2} = 0

What are the solutions to the equation -3d^2-4d+5=0 ?
The solutions to the equation -3d^2-4d+5=0 are d=-(2+sqrt(19))/3 ,d=(sqrt(19)-2)/3
Find the zeros of -3d^2-4d+5=0
The zeros of -3d^2-4d+5=0 are d=-(2+sqrt(19))/3 ,d=(sqrt(19)-2)/3