What are the solutions to the equation 3t^2-18t+24=0 ?

The solutions to the equation 3t^2-18t+24=0 are t=4,t=2

Find the zeros of 3t^2-18t+24=0

The zeros of 3t^2-18t+24=0 are t=4,t=2

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3 t^{2} - 18 t + 24 = 0

t = 4, t = 2

Solution steps

3 t^{2} - 18 t + 24 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 3 \cdot 24}{2 \cdot 3}

(- 18)^{2} - 4 \cdot 3 \cdot 24 = 6

t_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- ( - 18 ) + 6}{2 \cdot 3}, t_{2} = \frac{- ( - 18 ) - 6}{2 \cdot 3}

t = \frac{- ( - 18 ) + 6}{2 \cdot 3} : 4

t = \frac{- ( - 18 ) - 6}{2 \cdot 3} : 2

The solutions to the quadratic equation are:

t = 4, t = 2

3 2^{\frac{2}{5}}

5^{- \frac{3}{5}}

- 5 (w - 1) \leq 45 or 11 + w < 21

2 x^{2} + 13 x - 15 = 0

x + 7 = 2

What are the solutions to the equation 3t^2-18t+24=0 ?
The solutions to the equation 3t^2-18t+24=0 are t=4,t=2
Find the zeros of 3t^2-18t+24=0
The zeros of 3t^2-18t+24=0 are t=4,t=2