What are the solutions to the equation -3(t-1)^2+12=0 ?

The solutions to the equation -3(t-1)^2+12=0 are t=3,t=-1

Find the zeros of -3(t-1)^2+12=0

The zeros of -3(t-1)^2+12=0 are t=3,t=-1

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- 3 (t - 1)^{2} + 12 = 0

t = 3, t = - 1

Solution steps

- 3 (t - 1)^{2} + 12 = 0

Move

12

to the right side

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

Divide both sides by

- 3

(t - 1)^{2} = 4

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

t - 1 = 4 : t = 3

Solve

t - 1 = - 4 : t = - 1

The solutions to the quadratic equation are:

t = 3, t = - 1

- x^{2} + 9 = 25

x^{2} + 23 - 140 = 0

x, x y^{3} = - 3 x^{2} y^{4}

3 r^{2} = 18

m^{2} = 225

What are the solutions to the equation -3(t-1)^2+12=0 ?
The solutions to the equation -3(t-1)^2+12=0 are t=3,t=-1
Find the zeros of -3(t-1)^2+12=0
The zeros of -3(t-1)^2+12=0 are t=3,t=-1