What are the solutions to the equation s^2-4s+2=0 ?

The solutions to the equation s^2-4s+2=0 are s=2+sqrt(2),s=2-sqrt(2)

Find the zeros of s^2-4s+2=0

The zeros of s^2-4s+2=0 are s=2+sqrt(2),s=2-sqrt(2)

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s^{2} - 4 s + 2 = 0

s = 2 + 2, s = 2 - 2

Solution steps

s^{2} - 4 s + 2 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 2 = 22

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

Separate the solutions

s_{1} = \frac{- ( - 4 ) + 2 2}{2 \cdot 1}, s_{2} = \frac{- ( - 4 ) - 2 2}{2 \cdot 1}

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

s = \frac{- ( - 4 ) - 2 2}{2 \cdot 1} : 2 - 2

The solutions to the quadratic equation are:

s = 2 + 2, s = 2 - 2

49 x^{2} - 154 x + 121

a c + 9 a - 2 c - 18

\frac{7 ^{1}}{7 ^{3}} = 7

11 - j \leq 8

v^{2} + 20 v + 19

What are the solutions to the equation s^2-4s+2=0 ?
The solutions to the equation s^2-4s+2=0 are s=2+sqrt(2),s=2-sqrt(2)
Find the zeros of s^2-4s+2=0
The zeros of s^2-4s+2=0 are s=2+sqrt(2),s=2-sqrt(2)