What are the solutions to the equation 2x^2-20x+41=0 ?

The solutions to the equation 2x^2-20x+41=0 are x=(10+3sqrt(2))/2 ,x=(10-3sqrt(2))/2

Find the zeros of 2x^2-20x+41=0

The zeros of 2x^2-20x+41=0 are x=(10+3sqrt(2))/2 ,x=(10-3sqrt(2))/2

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2 x^{2} - 20 x + 41 = 0

x = \frac{10 + 3 2}{2}, x = \frac{10 - 3 2}{2}

Decimal

x = 7.12132 \dots, x = 2.87867 \dots

Solution steps

2 x^{2} - 20 x + 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 \cdot 41}{2 \cdot 2}

(- 20)^{2} - 4 \cdot 2 \cdot 41 = 62

x_{1, 2} = \frac{- ( - 20 ) \pm 6 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 6 2}{2 \cdot 2}, x_{2} = \frac{- ( - 20 ) - 6 2}{2 \cdot 2}

x = \frac{- ( - 20 ) + 6 2}{2 \cdot 2} : \frac{10 + 3 2}{2}

x = \frac{- ( - 20 ) - 6 2}{2 \cdot 2} : \frac{10 - 3 2}{2}

The solutions to the quadratic equation are:

x = \frac{10 + 3 2}{2}, x = \frac{10 - 3 2}{2}

a x^{2}

x^{2} + 4 x + 4

t, s = u t + \frac{1}{2} a t^{2}

n^{2} + 17 n + 22 = 7

a, c^{2} = a^{2} + b^{2}

What are the solutions to the equation 2x^2-20x+41=0 ?
The solutions to the equation 2x^2-20x+41=0 are x=(10+3sqrt(2))/2 ,x=(10-3sqrt(2))/2
Find the zeros of 2x^2-20x+41=0
The zeros of 2x^2-20x+41=0 are x=(10+3sqrt(2))/2 ,x=(10-3sqrt(2))/2