What are the solutions to the equation 9=t^2-2t+1 ?

The solutions to the equation 9=t^2-2t+1 are t=4,t=-2

Find the zeros of 9=t^2-2t+1

The zeros of 9=t^2-2t+1 are t=4,t=-2

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

t = 4, t = - 2

Solution steps

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

Switch sides

t^{2} - 2 t + 1 = 9

Move

9

to the left side

t^{2} - 2 t - 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 8) = 6

t_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 1}, t_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 1}

t = \frac{- ( - 2 ) + 6}{2 \cdot 1} : 4

t = \frac{- ( - 2 ) - 6}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

t = 4, t = - 2

5 - 3 (a + 6) = a - 1 + 8 a

\frac{32 a ^{5} b ^{- 3}}{8 a ^{- 2} b ^{6}}

6 x^{2} + 17 x = 3

\frac{5.0 5 ^{2} - 5 ( 5.05 )}{5.0 5 ^{2} - 25}

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

What are the solutions to the equation 9=t^2-2t+1 ?
The solutions to the equation 9=t^2-2t+1 are t=4,t=-2
Find the zeros of 9=t^2-2t+1
The zeros of 9=t^2-2t+1 are t=4,t=-2