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

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

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

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

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

a = 6, a = - 2

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

a = 6, a = - 2

\frac{10}{- 8}

\frac{2}{5} k - \frac{3}{5} + \frac{1}{10} k

14 x^{2} + x - 3

19 - 6 (- k - 2)

\frac{1 0 ^{2}}{1 0 ^{10}}

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