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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Decimal Notation

a = 0.78629 \dots, a = - 2.11963 \dots

Solution steps

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

Solve with the quadratic formula

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

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

a_{1, 2} = \frac{- 4 \pm 2 19}{2 \cdot 3}

Separate the solutions

a_{1} = \frac{- 4 + 2 19}{2 \cdot 3}, a_{2} = \frac{- 4 - 2 19}{2 \cdot 3}

a = \frac{- 4 + 2 19}{2 \cdot 3} : \frac{- 2 + 19}{3}

a = \frac{- 4 - 2 19}{2 \cdot 3} : - \frac{2 + 19}{3}

The solutions to the quadratic equation are:

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

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

b (w) = 4 w^{2} + 5 w + 100

\frac{x ^{2}}{16} = 1

x^{2} - 2 x + 28 = 0

y, y + y^{2} = t^{2} - 4

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