What are the solutions to the equation s^2-4s-12=0 ?

The solutions to the equation s^2-4s-12=0 are s=6,s=-2

Find the zeros of s^2-4s-12=0

The zeros of s^2-4s-12=0 are s=6,s=-2

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s^{2} - 4 s - 12 = 0

s = 6, s = - 2

Solution steps

s^{2} - 4 s - 12 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

s_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, s_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

s = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

s = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

s = 6, s = - 2

(\frac{3 + 41}{2}) x^{2} + (\frac{3 - 41}{2}) = 9

- 4 x > 12

- 2 (x + 3) > - 16

3 x - 4 = 13

\frac{r ^{2}}{100} - \frac{t ^{2}}{49}

What are the solutions to the equation s^2-4s-12=0 ?
The solutions to the equation s^2-4s-12=0 are s=6,s=-2
Find the zeros of s^2-4s-12=0
The zeros of s^2-4s-12=0 are s=6,s=-2