What are the solutions to the equation (2t-3)^2-4(2t-3)^2+3=0 ?

The solutions to the equation (2t-3)^2-4(2t-3)^2+3=0 are t=2,t=1

Find the zeros of (2t-3)^2-4(2t-3)^2+3=0

The zeros of (2t-3)^2-4(2t-3)^2+3=0 are t=2,t=1

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(2 t - 3)^{2} - 4 (2 t - 3)^{2} + 3 = 0

t = 2, t = 1

Solution steps

(2 t - 3)^{2} - 4 (2 t - 3)^{2} + 3 = 0

Move

3

to the right side

(2 t - 3)^{2} - 4 (2 t - 3)^{2} = - 3

Add similar elements:

(2 t - 3)^{2} - 4 (2 t - 3)^{2} = - 3 (2 t - 3)^{2}

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

Divide both sides by

- 3

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

For

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

the solutions are

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

Solve

2 t - 3 = 1 : t = 2

Solve

2 t - 3 = - 1 : t = 1

The solutions to the quadratic equation are:

t = 2, t = 1

2 x^{2} + 4 x + 7 = 0

\frac{2}{3} x = 2

- 8 < 2 (x + 4)

13 x + 25 < - 1

8 x^{2} - 14 x - 9

What are the solutions to the equation (2t-3)^2-4(2t-3)^2+3=0 ?
The solutions to the equation (2t-3)^2-4(2t-3)^2+3=0 are t=2,t=1
Find the zeros of (2t-3)^2-4(2t-3)^2+3=0
The zeros of (2t-3)^2-4(2t-3)^2+3=0 are t=2,t=1