What are the solutions to the equation t^2-2t+1=36 ?

The solutions to the equation t^2-2t+1=36 are t=7,t=-5

Find the zeros of t^2-2t+1=36

The zeros of t^2-2t+1=36 are t=7,t=-5

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t^{2} - 2 t + 1 = 36

t = 7, t = - 5

Solution steps

t^{2} - 2 t + 1 = 36

Move

36

to the left side

t^{2} - 2 t - 35 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 35 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 35) = 12

t_{1, 2} = \frac{- ( - 2 ) \pm 12}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 2 ) + 12}{2 \cdot 1}, t_{2} = \frac{- ( - 2 ) - 12}{2 \cdot 1}

t = \frac{- ( - 2 ) + 12}{2 \cdot 1} : 7

t = \frac{- ( - 2 ) - 12}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

t = 7, t = - 5

x, x^{2} + y^{2} + 6 x - 4 y + 13 = 0

3 t^{2} - t - 6

4 x^{2} + 1.32 x - 0.33 = 0

x, x^{2} - 121 = 0

x, g = a (x - h)^{2} + k

What are the solutions to the equation t^2-2t+1=36 ?
The solutions to the equation t^2-2t+1=36 are t=7,t=-5
Find the zeros of t^2-2t+1=36
The zeros of t^2-2t+1=36 are t=7,t=-5