What are the solutions to the equation 3t^2-12t+9=0 ?

The solutions to the equation 3t^2-12t+9=0 are t=3,t=1

Find the zeros of 3t^2-12t+9=0

The zeros of 3t^2-12t+9=0 are t=3,t=1

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3 t^{2} - 12 t + 9 = 0

t = 3, t = 1

Solution steps

3 t^{2} - 12 t + 9 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 9}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 9 = 6

t_{1, 2} = \frac{- ( - 12 ) \pm 6}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- ( - 12 ) + 6}{2 \cdot 3}, t_{2} = \frac{- ( - 12 ) - 6}{2 \cdot 3}

t = \frac{- ( - 12 ) + 6}{2 \cdot 3} : 3

t = \frac{- ( - 12 ) - 6}{2 \cdot 3} : 1

The solutions to the quadratic equation are:

t = 3, t = 1

\frac{b ^{- 2}}{b ^{- 3}}

(x - 8) (x - 2) (x + 4) > 0

23 x + 7 = 28 x + 57

- 4 < - 2 x \leq 12

(8 - 5 x) (x + 2) + 4 (x - 2) (x + 1) + 2 (x - 2) (x + 2) = 0

What are the solutions to the equation 3t^2-12t+9=0 ?
The solutions to the equation 3t^2-12t+9=0 are t=3,t=1
Find the zeros of 3t^2-12t+9=0
The zeros of 3t^2-12t+9=0 are t=3,t=1