What are the solutions to the equation s^2-2s+10=0 ?

The solutions to the equation s^2-2s+10=0 are s=1+3i,s=1-3i

Find the zeros of s^2-2s+10=0

The zeros of s^2-2s+10=0 are s=1+3i,s=1-3i

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s^{2} - 2 s + 10 = 0

s = 1 + 3 i, s = 1 - 3 i

Solution steps

s^{2} - 2 s + 10 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 10 : 6 i

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

Separate the solutions

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

s = \frac{- ( - 2 ) + 6 i}{2 \cdot 1} : 1 + 3 i

s = \frac{- ( - 2 ) - 6 i}{2 \cdot 1} : 1 - 3 i

The solutions to the quadratic equation are:

s = 1 + 3 i, s = 1 - 3 i

\frac{3}{4} \times 4

10 - 25 x = 5

\frac{1}{3} x + \frac{2}{5} = \frac{1}{5} x - \frac{2}{5}

3 (2 - x) + 4 < 16

- 2 k + (- 4 k) + 5

What are the solutions to the equation s^2-2s+10=0 ?
The solutions to the equation s^2-2s+10=0 are s=1+3i,s=1-3i
Find the zeros of s^2-2s+10=0
The zeros of s^2-2s+10=0 are s=1+3i,s=1-3i