What are the solutions to the equation 3x^2+5=41 ?

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

Find the zeros of 3x^2+5=41

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

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3 x^{2} + 5 = 41

x = 23, x = - 23

Decimal Notation

x = 3.46410 \dots, x = - 3.46410 \dots

Solution steps

3 x^{2} + 5 = 41

Move

5

to the right side

3 x^{2} = 36

Divide both sides by

3

x^{2} = 12

For

x^{2} = f (a)

the solutions are

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

x = 12, x = - 12

12 = 23

- 12 = - 23

x = 23, x = - 23

x, x^{2} - x = 0

6 x^{2} + 2 x + 6 = 0

x^{2} - 12 x + 11 = 0

25 \cdot 1 0^{- 9} s^{2} + 0.35 \cdot 1 0^{- 3} s + 1 = 0

4 - 4 x + x^{2} = 0

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