What are the solutions to the equation 3a^2-12a+12=0 ?

The solutions to the equation 3a^2-12a+12=0 are a=2

Find the zeros of 3a^2-12a+12=0

The zeros of 3a^2-12a+12=0 are a=2

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

a = 2

Solution steps

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

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 12}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 12 = 0

a_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 3}

a = \frac{- ( - 12 )}{2 \cdot 3}

\frac{- ( - 12 )}{2 \cdot 3} = 2

a = 2

The solution to the quadratic equation is:

a = 2

- 10 + k < - 9 + 2 k < 4 + k

- 6 (3 k + 5) = 5 k + 39

∣ 2 x + 1 ∣ \geq 5

13 x < - 3 or 2 x > 0

\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}

What are the solutions to the equation 3a^2-12a+12=0 ?
The solutions to the equation 3a^2-12a+12=0 are a=2
Find the zeros of 3a^2-12a+12=0
The zeros of 3a^2-12a+12=0 are a=2