What are the solutions to the equation s^2-3s+2=0 ?

The solutions to the equation s^2-3s+2=0 are s=2,s=1

Find the zeros of s^2-3s+2=0

The zeros of s^2-3s+2=0 are s=2,s=1

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s^{2} - 3 s + 2 = 0

s = 2, s = 1

Solution steps

s^{2} - 3 s + 2 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 2 = 1

s_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, s_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

s = \frac{- ( - 3 ) + 1}{2 \cdot 1} : 2

s = \frac{- ( - 3 ) - 1}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

s = 2, s = 1

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- 9 x + 2 > 18

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20 x + 5 > 21 x + 15

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