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

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

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

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

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a^{2} - 12 = - 4 a

a = 2, a = - 6

Solution steps

a^{2} - 12 = - 4 a

Move

4 a

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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} : 2

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

The solutions to the quadratic equation are:

a = 2, a = - 6

3 = 51 - 2 x^{2}

x^{2} + 7 = 4 (x + 7)

x^{2} + 10 x + 21 = 0

6 m^{2} + 7 m - 3 = 0

- 2 x^{2} - x + 4 = 0

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