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

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

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

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

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2 x^{2} - 41 = 0

x = \frac{41}{2}, x = - \frac{41}{2}

Decimal Notation

x = 4.52769 \dots, x = - 4.52769 \dots

Solution steps

2 x^{2} - 41 = 0

Move

41

to the right side

2 x^{2} = 41

Divide both sides by

2

x^{2} = \frac{41}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = \frac{41}{2}, x = - \frac{41}{2}

x^{2} - 6 x + 9 - y^{2}

12 - (u + 7) < 3 (u - 1)

(n^{6} k^{14})^{5}

(\frac{25}{64})^{- \frac{3}{2}}

x y + 3 x + 2 y + 6

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