What are the solutions to the equation 4x^2-5=x^2+7 ?

The solutions to the equation 4x^2-5=x^2+7 are x=2,x=-2

Find the zeros of 4x^2-5=x^2+7

The zeros of 4x^2-5=x^2+7 are x=2,x=-2

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4 x^{2} - 5 = x^{2} + 7

x = 2, x = - 2

Solution steps

4 x^{2} - 5 = x^{2} + 7

Move

5

to the right side

4 x^{2} = x^{2} + 12

Move

x^{2}

to the left side

3 x^{2} = 12

Divide both sides by

3

x^{2} = 4

For

x^{2} = f (a)

the solutions are

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

x = 4, x = - 4

4 = 2

- 4 = - 2

x = 2, x = - 2

x, (x + y - 2)^{2} = e^{y} - 1

(x - 4)^{2} + (2 x - 15 + 2)^{2} = 5

- 4.91 x^{2} + 2 = 0

x^{2} + (x + 1)^{2} + (x + 2)^{2} = 110

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

What are the solutions to the equation 4x^2-5=x^2+7 ?
The solutions to the equation 4x^2-5=x^2+7 are x=2,x=-2
Find the zeros of 4x^2-5=x^2+7
The zeros of 4x^2-5=x^2+7 are x=2,x=-2