What are the solutions to the equation quadraticformula 3x^2-5x-1=0 ?

The solutions to the equation quadraticformula 3x^2-5x-1=0 are x=(5+sqrt(37))/6 ,x=(5-sqrt(37))/6

Find the zeros of quadraticformula 3x^2-5x-1=0

The zeros of quadraticformula 3x^2-5x-1=0 are x=(5+sqrt(37))/6 ,x=(5-sqrt(37))/6

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3 x^{2} - 5 x - 1 = 0

x = \frac{5 + 37}{6}, x = \frac{5 - 37}{6}

Decimal Notation

x = 1.84712 \dots, x = - 0.18046 \dots

Solution steps

3 x^{2} - 5 x - 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 3 (- 1) = 37

x_{1, 2} = \frac{- ( - 5 ) \pm 37}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 37}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 37}{2 \cdot 3}

x = \frac{- ( - 5 ) + 37}{2 \cdot 3} : \frac{5 + 37}{6}

x = \frac{- ( - 5 ) - 37}{2 \cdot 3} : \frac{5 - 37}{6}

The solutions to the quadratic equation are:

x = \frac{5 + 37}{6}, x = \frac{5 - 37}{6}

- 3 x^{2} - 14 x + 5 = 0

7 (x + 18)^{2} = 2800

(\frac{y ^{2}}{2}) (7.6) = 2.01 (17.5 - y) + 2.01 (10.5 - y)

0 = - 16 x^{2} + 64

- 25 b^{2} = 0

What are the solutions to the equation quadraticformula 3x^2-5x-1=0 ?
The solutions to the equation quadraticformula 3x^2-5x-1=0 are x=(5+sqrt(37))/6 ,x=(5-sqrt(37))/6
Find the zeros of quadraticformula 3x^2-5x-1=0
The zeros of quadraticformula 3x^2-5x-1=0 are x=(5+sqrt(37))/6 ,x=(5-sqrt(37))/6