What are the solutions to the equation 6t^2+5=2t^2+1 ?

The solutions to the equation 6t^2+5=2t^2+1 are t=i,t=-i

Find the zeros of 6t^2+5=2t^2+1

The zeros of 6t^2+5=2t^2+1 are t=i,t=-i

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6 t^{2} + 5 = 2 t^{2} + 1

t = i, t = - i

Solution steps

6 t^{2} + 5 = 2 t^{2} + 1

Move

5

to the right side

6 t^{2} = 2 t^{2} - 4

Move

2 t^{2}

to the left side

4 t^{2} = - 4

Divide both sides by

4

t^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

t = - 1, t = - - 1

- 1 = i

- - 1 = - i

t = i, t = - i

∣ 4 - x ∣ < ∣ 3 + 2 x ∣

∣ x + 19 ∣ < 5

b^{4} (3 a b^{3}) (2 a^{2} b^{- 5})

119 - 0.16 \cdot 28 \cdot 9.81 = 28 \cdot x

x^{2} + (6 x + 22)^{2} = (6 x + 23)^{2}

What are the solutions to the equation 6t^2+5=2t^2+1 ?
The solutions to the equation 6t^2+5=2t^2+1 are t=i,t=-i
Find the zeros of 6t^2+5=2t^2+1
The zeros of 6t^2+5=2t^2+1 are t=i,t=-i