What are the solutions to the equation s^2+7s=-20 ?

The solutions to the equation s^2+7s=-20 are s=-7/2+i(sqrt(31))/2 ,s=-7/2-i(sqrt(31))/2

Find the zeros of s^2+7s=-20

The zeros of s^2+7s=-20 are s=-7/2+i(sqrt(31))/2 ,s=-7/2-i(sqrt(31))/2

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s^{2} + 7 s = - 20

s = - \frac{7}{2} + i \frac{31}{2}, s = - \frac{7}{2} - i \frac{31}{2}

Solution steps

s^{2} + 7 s = - 20

Move

20

to the left side

s^{2} + 7 s + 20 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

Simplify

7^{2} - 4 \cdot 1 \cdot 20 : 31 i

s_{1, 2} = \frac{- 7 \pm 31 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 7 + 31 i}{2 \cdot 1}, s_{2} = \frac{- 7 - 31 i}{2 \cdot 1}

s = \frac{- 7 + 31 i}{2 \cdot 1} : - \frac{7}{2} + i \frac{31}{2}

s = \frac{- 7 - 31 i}{2 \cdot 1} : - \frac{7}{2} - i \frac{31}{2}

The solutions to the quadratic equation are:

s = - \frac{7}{2} + i \frac{31}{2}, s = - \frac{7}{2} - i \frac{31}{2}

lo g_{7} (7^{6})

5 < \frac{13 x - 5}{4} < 8

2 x^{3} + 2

4 g > 8

∣ x + 6 ∣ = 2 x - 1

What are the solutions to the equation s^2+7s=-20 ?
The solutions to the equation s^2+7s=-20 are s=-7/2+i(sqrt(31))/2 ,s=-7/2-i(sqrt(31))/2
Find the zeros of s^2+7s=-20
The zeros of s^2+7s=-20 are s=-7/2+i(sqrt(31))/2 ,s=-7/2-i(sqrt(31))/2