What are the solutions to the equation s^2+2s-6=0 ?

The solutions to the equation s^2+2s-6=0 are s=-1+sqrt(7),s=-1-sqrt(7)

Find the zeros of s^2+2s-6=0

The zeros of s^2+2s-6=0 are s=-1+sqrt(7),s=-1-sqrt(7)

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s^{2} + 2 s - 6 = 0

s = - 1 + 7, s = - 1 - 7

Solution steps

s^{2} + 2 s - 6 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 6) = 27

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

s = - 1 + 7, s = - 1 - 7

0.9 x^{2} - x + 0.9 = 0

(7^{2})^{- 7}

120 = 495 - \frac{7}{5} q

5 x^{2} - 44 x + 120 = - 30 + 11 x

(\frac{6 l m ^{2}}{3 l ^{3} m ^{6}})^{2}

What are the solutions to the equation s^2+2s-6=0 ?
The solutions to the equation s^2+2s-6=0 are s=-1+sqrt(7),s=-1-sqrt(7)
Find the zeros of s^2+2s-6=0
The zeros of s^2+2s-6=0 are s=-1+sqrt(7),s=-1-sqrt(7)