What are the solutions to the equation s^2+40s+80=0 ?

The solutions to the equation s^2+40s+80=0 are s=4(2sqrt(5)-5),s=-4(5+2sqrt(5))

Find the zeros of s^2+40s+80=0

The zeros of s^2+40s+80=0 are s=4(2sqrt(5)-5),s=-4(5+2sqrt(5))

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s^{2} + 40 s + 80 = 0

s = 4 (25 - 5), s = - 4 (5 + 25)

Decimal Notation

s = - 2.11145 \dots, s = - 37.88854 \dots

Solution steps

s^{2} + 40 s + 80 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 1 \cdot 80}{2 \cdot 1}

4 0^{2} - 4 \cdot 1 \cdot 80 = 165

s_{1, 2} = \frac{- 40 \pm 16 5}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 40 + 16 5}{2 \cdot 1}, s_{2} = \frac{- 40 - 16 5}{2 \cdot 1}

s = \frac{- 40 + 16 5}{2 \cdot 1} : 4 (25 - 5)

s = \frac{- 40 - 16 5}{2 \cdot 1} : - 4 (5 + 25)

The solutions to the quadratic equation are:

s = 4 (25 - 5), s = - 4 (5 + 25)

5 x^{2} - 11 x - 168 = 0

- 16 t^{2} + 32 t + 2 = 0

x^{2} - 10 x + 30

0 = 9 x^{2} - 49

4 x + 16 = 2 x^{2} + 10

What are the solutions to the equation s^2+40s+80=0 ?
The solutions to the equation s^2+40s+80=0 are s=4(2sqrt(5)-5),s=-4(5+2sqrt(5))
Find the zeros of s^2+40s+80=0
The zeros of s^2+40s+80=0 are s=4(2sqrt(5)-5),s=-4(5+2sqrt(5))