What are the solutions to the equation 6t^2-18t+12=0 ?

The solutions to the equation 6t^2-18t+12=0 are t=2,t=1

Find the zeros of 6t^2-18t+12=0

The zeros of 6t^2-18t+12=0 are t=2,t=1

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6 t^{2} - 18 t + 12 = 0

t = 2, t = 1

Solution steps

6 t^{2} - 18 t + 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 6 \cdot 12}{2 \cdot 6}

(- 18)^{2} - 4 \cdot 6 \cdot 12 = 6

t_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 \cdot 6}

Separate the solutions

t_{1} = \frac{- ( - 18 ) + 6}{2 \cdot 6}, t_{2} = \frac{- ( - 18 ) - 6}{2 \cdot 6}

t = \frac{- ( - 18 ) + 6}{2 \cdot 6} : 2

t = \frac{- ( - 18 ) - 6}{2 \cdot 6} : 1

The solutions to the quadratic equation are:

t = 2, t = 1

3 x^{2} + 11 x - 10 = 0

\frac{1}{3} (x + 5)^{2} = 7

x, x^{2} + y^{2} = e^{- z}

2 x^{2} - 1 = 2 x + 3

x^{2} - x + \frac{1}{6} = 0

What are the solutions to the equation 6t^2-18t+12=0 ?
The solutions to the equation 6t^2-18t+12=0 are t=2,t=1
Find the zeros of 6t^2-18t+12=0
The zeros of 6t^2-18t+12=0 are t=2,t=1