What are the solutions to the equation 2x^2+5=41 ?

The solutions to the equation 2x^2+5=41 are x=3sqrt(2),x=-3sqrt(2)

Find the zeros of 2x^2+5=41

The zeros of 2x^2+5=41 are x=3sqrt(2),x=-3sqrt(2)

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2 x^{2} + 5 = 41

x = 32, x = - 32

Solution steps

2 x^{2} + 5 = 41

Move

5

to the right side

2 x^{2} = 36

Divide both sides by

2

x^{2} = 18

For

x^{2} = f (a)

the solutions are

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

x = 18, x = - 18

18 = 32

- 18 = - 32

x = 32, x = - 32

\frac{6 x + 1}{x - 2} \leq 0

\frac{3}{5} y = 6

4 (x + 2) > 12

(x^{2} + x y^{2})^{2} - (x y + y^{2})^{2}

(x^{- 2})^{2}

What are the solutions to the equation 2x^2+5=41 ?
The solutions to the equation 2x^2+5=41 are x=3sqrt(2),x=-3sqrt(2)
Find the zeros of 2x^2+5=41
The zeros of 2x^2+5=41 are x=3sqrt(2),x=-3sqrt(2)