What are the solutions to the equation 2x^2-3=-x^2 ?

The solutions to the equation 2x^2-3=-x^2 are x=1,x=-1

Find the zeros of 2x^2-3=-x^2

The zeros of 2x^2-3=-x^2 are x=1,x=-1

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2 x^{2} - 3 = - x^{2}

x = 1, x = - 1

Solution steps

2 x^{2} - 3 = - x^{2}

Move

3

to the right side

2 x^{2} = - x^{2} + 3

Move

x^{2}

to the left side

3 x^{2} = 3

Divide both sides by

3

x^{2} = 1

For

x^{2} = f (a)

the solutions are

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

x = 1, x = - 1

1 = 1

- 1 = - 1

x = 1, x = - 1

3 x^{2} + 7 x = 2

409 6^{\frac{1}{6}}

- 3 n (2 n - 7) = 0

∣ x - 3 ∣ \geq 5

25 x^{2} y^{2} - 4 x^{2} y + 16 x^{2}

What are the solutions to the equation 2x^2-3=-x^2 ?
The solutions to the equation 2x^2-3=-x^2 are x=1,x=-1
Find the zeros of 2x^2-3=-x^2
The zeros of 2x^2-3=-x^2 are x=1,x=-1