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

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

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

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

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3 x^{2} - 4 x - 5 = 0

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

Decimal Notation

x = 2.11963 \dots, x = - 0.78629 \dots

Solution steps

3 x^{2} - 4 x - 5 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

6 y^{2} - 5 y - 21

\frac{5}{4 - i}

y^{2} - 15 y + 36

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

\frac{1}{2} - (- \frac{1}{2})

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