What are the solutions to the equation roots s^2+1.2s+1 ?

The solutions to the equation roots s^2+1.2s+1 are s=-3/5+i 4/5 ,s=-3/5-i 4/5

Find the zeros of roots s^2+1.2s+1

The zeros of roots s^2+1.2s+1 are s=-3/5+i 4/5 ,s=-3/5-i 4/5

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roots

s^{2} + 1.2 s + 1

s = - \frac{3}{5} + i \frac{4}{5}, s = - \frac{3}{5} - i \frac{4}{5}

Solution steps

s^{2} + 1.2 s + 1

The roots are the intercepts with the x-axis

(y = 0)

s^{2} + 1.2 s + 1 = 0

Multiply both sides by

10

10 s^{2} + 12 s + 10 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 10 \cdot 10}{2 \cdot 10}

Simplify

1 2^{2} - 4 \cdot 10 \cdot 10 : 16 i

s_{1, 2} = \frac{- 12 \pm 16 i}{2 \cdot 10}

Separate the solutions

s_{1} = \frac{- 12 + 16 i}{2 \cdot 10}, s_{2} = \frac{- 12 - 16 i}{2 \cdot 10}

s = \frac{- 12 + 16 i}{2 \cdot 10} : - \frac{3}{5} + i \frac{4}{5}

s = \frac{- 12 - 16 i}{2 \cdot 10} : - \frac{3}{5} - i \frac{4}{5}

The solutions to the quadratic equation are:

s = - \frac{3}{5} + i \frac{4}{5}, s = - \frac{3}{5} - i \frac{4}{5}

a, a^{2} - 3 a y^{2} + 3 y^{4} - b = 0

- 25 x^{2} + 20 x - 4 = 0

3 (x^{2} - 7) - 4 = - 7

y, y^{2} = x^{3} - x

4 x^{2} + 12 x - 5

What are the solutions to the equation roots s^2+1.2s+1 ?
The solutions to the equation roots s^2+1.2s+1 are s=-3/5+i 4/5 ,s=-3/5-i 4/5
Find the zeros of roots s^2+1.2s+1
The zeros of roots s^2+1.2s+1 are s=-3/5+i 4/5 ,s=-3/5-i 4/5