What are the solutions to the equation (2/3)(t^2-1)=12 ?

The solutions to the equation (2/3)(t^2-1)=12 are t=sqrt(19),t=-sqrt(19)

Find the zeros of (2/3)(t^2-1)=12

The zeros of (2/3)(t^2-1)=12 are t=sqrt(19),t=-sqrt(19)

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(\frac{2}{3}) (t^{2} - 1) = 12

t = 19, t = - 19

Decimal Notation

t = 4.35889 \dots, t = - 4.35889 \dots

Solution steps

(\frac{2}{3}) (t^{2} - 1) = 12

Multiply both sides by

3

2 (t^{2} - 1) = 36

Divide both sides by

2

t^{2} - 1 = 18

Move

1

to the right side

t^{2} = 19

For

x^{2} = f (a)

the solutions are

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

t = 19, t = - 19

\frac{x}{3} - 1 + \frac{x ^{2}}{2} = - 2 x^{2} + x - \frac{7}{6}

6 a^{2} + 5 a - 4 = 2

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

(3 x + 4)^{2} = 8

m (m + 2) = - 3 - 2 m

What are the solutions to the equation (2/3)(t^2-1)=12 ?
The solutions to the equation (2/3)(t^2-1)=12 are t=sqrt(19),t=-sqrt(19)
Find the zeros of (2/3)(t^2-1)=12
The zeros of (2/3)(t^2-1)=12 are t=sqrt(19),t=-sqrt(19)