What are the solutions to the equation 6x^2-9x+1=0 ?

The solutions to the equation 6x^2-9x+1=0 are x=(9+sqrt(57))/(12),x=(9-sqrt(57))/(12)

Find the zeros of 6x^2-9x+1=0

The zeros of 6x^2-9x+1=0 are x=(9+sqrt(57))/(12),x=(9-sqrt(57))/(12)

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6 x^{2} - 9 x + 1 = 0

x = \frac{9 + 57}{12}, x = \frac{9 - 57}{12}

Decimal Notation

x = 1.37915 \dots, x = 0.12084 \dots

Solution steps

6 x^{2} - 9 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

(- 9)^{2} - 4 \cdot 6 \cdot 1 = 57

x_{1, 2} = \frac{- ( - 9 ) \pm 57}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 57}{2 \cdot 6}, x_{2} = \frac{- ( - 9 ) - 57}{2 \cdot 6}

x = \frac{- ( - 9 ) + 57}{2 \cdot 6} : \frac{9 + 57}{12}

x = \frac{- ( - 9 ) - 57}{2 \cdot 6} : \frac{9 - 57}{12}

The solutions to the quadratic equation are:

x = \frac{9 + 57}{12}, x = \frac{9 - 57}{12}

3 = \frac{3}{2} (7 x + 10)

x^{2} - 9 x y + 14 y^{2}

- 1 \leq \frac{3}{4} x - 4 < 5

a^{2} - 9 a

What are the solutions to the equation 6x^2-9x+1=0 ?
The solutions to the equation 6x^2-9x+1=0 are x=(9+sqrt(57))/(12),x=(9-sqrt(57))/(12)
Find the zeros of 6x^2-9x+1=0
The zeros of 6x^2-9x+1=0 are x=(9+sqrt(57))/(12),x=(9-sqrt(57))/(12)