What are the solutions to the equation 2x^2-18x+1=0 ?

The solutions to the equation 2x^2-18x+1=0 are x=(9+sqrt(79))/2 ,x=(9-sqrt(79))/2

Find the zeros of 2x^2-18x+1=0

The zeros of 2x^2-18x+1=0 are x=(9+sqrt(79))/2 ,x=(9-sqrt(79))/2

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2 x^{2} - 18 x + 1 = 0

x = \frac{9 + 79}{2}, x = \frac{9 - 79}{2}

Decimal Notation

x = 8.94409 \dots, x = 0.05590 \dots

Solution steps

2 x^{2} - 18 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 18)^{2} - 4 \cdot 2 \cdot 1 = 279

x_{1, 2} = \frac{- ( - 18 ) \pm 2 79}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 2 79}{2 \cdot 2}, x_{2} = \frac{- ( - 18 ) - 2 79}{2 \cdot 2}

x = \frac{- ( - 18 ) + 2 79}{2 \cdot 2} : \frac{9 + 79}{2}

x = \frac{- ( - 18 ) - 2 79}{2 \cdot 2} : \frac{9 - 79}{2}

The solutions to the quadratic equation are:

x = \frac{9 + 79}{2}, x = \frac{9 - 79}{2}

v^{2} - 5 v - 14 = 0

- x^{2} + 8 x = 7

120 x^{2} + x - 75 = 0

\frac{1}{3} x^{2} - 5 x + 25 = 0

(6 + y)^{2} + y^{2} = 218

What are the solutions to the equation 2x^2-18x+1=0 ?
The solutions to the equation 2x^2-18x+1=0 are x=(9+sqrt(79))/2 ,x=(9-sqrt(79))/2
Find the zeros of 2x^2-18x+1=0
The zeros of 2x^2-18x+1=0 are x=(9+sqrt(79))/2 ,x=(9-sqrt(79))/2