What is the answer to (t-1)^2=(t+1)^2+16 ?

The answer to (t-1)^2=(t+1)^2+16 is t=-4

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(t-1)^2=(t+1)^2+16

(t - 1)^{2} = (t + 1)^{2} + 16

t = - 4

Solution steps

(t - 1)^{2} = (t + 1)^{2} + 16

Expand

(t - 1)^{2} : t^{2} - 2 t + 1

Expand

(t + 1)^{2} + 16 : t^{2} + 2 t + 17

t^{2} - 2 t + 1 = t^{2} + 2 t + 17

Move

1

to the right side

t^{2} - 2 t = t^{2} + 2 t + 16

Subtract

t^{2} + 2 t

from both sides

t^{2} - 2 t - (t^{2} + 2 t) = t^{2} + 2 t + 16 - (t^{2} + 2 t)

Simplify

- 4 t = 16

Divide both sides by

- 4

t = - 4

15 x^{2} + 14 x y - 16 y^{2}

∣ 3 b ∣ + 4 = 5

\frac{4 !}{2 ! 2 !}

\frac{8 g ^{2} - 12 g h}{2 g}

4 x^{2} - 8 x - 32 = 0

What is the answer to (t-1)^2=(t+1)^2+16 ?
The answer to (t-1)^2=(t+1)^2+16 is t=-4
Solve for t: (t-1)^2=(t+1)^2+16
The solution is t=-4