What are the solutions to the equation 2x^2-7x+1=0 ?

The solutions to the equation 2x^2-7x+1=0 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4

Find the zeros of 2x^2-7x+1=0

The zeros of 2x^2-7x+1=0 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4

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2 x^{2} - 7 x + 1 = 0

x = \frac{7 + 41}{4}, x = \frac{7 - 41}{4}

Decimal Notation

x = 3.35078 \dots, x = 0.14921 \dots

Solution steps

2 x^{2} - 7 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 1 = 41

x_{1, 2} = \frac{- ( - 7 ) \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 41}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 41}{2 \cdot 2}

x = \frac{- ( - 7 ) + 41}{2 \cdot 2} : \frac{7 + 41}{4}

x = \frac{- ( - 7 ) - 41}{2 \cdot 2} : \frac{7 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 41}{4}, x = \frac{7 - 41}{4}

lo g_{10} (3 x) < lo g_{10} (7 x - 8)

\frac{1}{4} + \frac{1}{20}

6 q^{2} - 3 q = 5 q^{2} - 8 q + 24

m^{2} + 6 m + 9 = 0

x + 1 + 4 \leq 9

What are the solutions to the equation 2x^2-7x+1=0 ?
The solutions to the equation 2x^2-7x+1=0 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4
Find the zeros of 2x^2-7x+1=0
The zeros of 2x^2-7x+1=0 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4