What are the solutions to the equation 3x^2-9x-5=0 ?

The solutions to the equation 3x^2-9x-5=0 are x=(9+sqrt(141))/6 ,x=(9-sqrt(141))/6

Find the zeros of 3x^2-9x-5=0

The zeros of 3x^2-9x-5=0 are x=(9+sqrt(141))/6 ,x=(9-sqrt(141))/6

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3 x^{2} - 9 x - 5 = 0

x = \frac{9 + 141}{6}, x = \frac{9 - 141}{6}

Decimal

x = 3.47905 \dots, x = - 0.47905 \dots

Solution steps

3 x^{2} - 9 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 3 ( - 5 )}{2 \cdot 3}

(- 9)^{2} - 4 \cdot 3 (- 5) = 141

x_{1, 2} = \frac{- ( - 9 ) \pm 141}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 141}{2 \cdot 3}, x_{2} = \frac{- ( - 9 ) - 141}{2 \cdot 3}

x = \frac{- ( - 9 ) + 141}{2 \cdot 3} : \frac{9 + 141}{6}

x = \frac{- ( - 9 ) - 141}{2 \cdot 3} : \frac{9 - 141}{6}

The solutions to the quadratic equation are:

x = \frac{9 + 141}{6}, x = \frac{9 - 141}{6}

w^{2} - 35 w = 0

64.77 + 25.96 T + 0.02134 T^{2} = 0

- 5 x^{2} - 5 x + 10 = 0

(5 - x) (- 1 - x) + 8 = 0

12 t + 9 = 5 t^{2}

What are the solutions to the equation 3x^2-9x-5=0 ?
The solutions to the equation 3x^2-9x-5=0 are x=(9+sqrt(141))/6 ,x=(9-sqrt(141))/6
Find the zeros of 3x^2-9x-5=0
The zeros of 3x^2-9x-5=0 are x=(9+sqrt(141))/6 ,x=(9-sqrt(141))/6