What are the solutions to the equation 3t^2-42t+144=0 ?

The solutions to the equation 3t^2-42t+144=0 are t=8,t=6

Find the zeros of 3t^2-42t+144=0

The zeros of 3t^2-42t+144=0 are t=8,t=6

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3 t^{2} - 42 t + 144 = 0

t = 8, t = 6

Solution steps

3 t^{2} - 42 t + 144 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 3 \cdot 144}{2 \cdot 3}

(- 42)^{2} - 4 \cdot 3 \cdot 144 = 6

t_{1, 2} = \frac{- ( - 42 ) \pm 6}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- ( - 42 ) + 6}{2 \cdot 3}, t_{2} = \frac{- ( - 42 ) - 6}{2 \cdot 3}

t = \frac{- ( - 42 ) + 6}{2 \cdot 3} : 8

t = \frac{- ( - 42 ) - 6}{2 \cdot 3} : 6

The solutions to the quadratic equation are:

t = 8, t = 6

\frac{3}{6} \cdot \frac{1}{2}

15 \cdot \frac{1}{6}

- \frac{3}{4} - \frac{2}{5}

x^{3} - 2 x^{2} + 10 x

∣ 8 - 2 x ∣ > 2

What are the solutions to the equation 3t^2-42t+144=0 ?
The solutions to the equation 3t^2-42t+144=0 are t=8,t=6
Find the zeros of 3t^2-42t+144=0
The zeros of 3t^2-42t+144=0 are t=8,t=6