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

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

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

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

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

s = 7, s = - 5

Solution steps

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

Move

36

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

s = 7, s = - 5

6 m^{2} + 3 m = - 3

- 2 x^{2} + 9 x - 12 = 0

2 (x - 2) (x - 3) = (x + 1) (x)

x^{2} - 6 x + 3^{2} + 16 = 0

12 x^{2} - 400 x + 2400 = 0

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