What are the solutions to the equation 4x^2+5=29 ?

The solutions to the equation 4x^2+5=29 are x=sqrt(6),x=-sqrt(6)

Find the zeros of 4x^2+5=29

The zeros of 4x^2+5=29 are x=sqrt(6),x=-sqrt(6)

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4 x^{2} + 5 = 29

x = 6, x = - 6

Decimal Notation

x = 2.44948 \dots, x = - 2.44948 \dots

Solution steps

4 x^{2} + 5 = 29

Move

5

to the right side

4 x^{2} = 24

Divide both sides by

4

x^{2} = 6

For

x^{2} = f (a)

the solutions are

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

x = 6, x = - 6

- 3 x^{2} - 2 = 0

256 = n^{2}

(x + 1)^{2} + (x - 1)^{2} = 0

x^{2} - 5 x - 6

x^{2} - 5 x - 3

What are the solutions to the equation 4x^2+5=29 ?
The solutions to the equation 4x^2+5=29 are x=sqrt(6),x=-sqrt(6)
Find the zeros of 4x^2+5=29
The zeros of 4x^2+5=29 are x=sqrt(6),x=-sqrt(6)