What are the solutions to the equation 3t^2-40t+128=0 ?

The solutions to the equation 3t^2-40t+128=0 are t=8,t= 16/3

Find the zeros of 3t^2-40t+128=0

The zeros of 3t^2-40t+128=0 are t=8,t= 16/3

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3 t^{2} - 40 t + 128 = 0

t = 8, t = \frac{16}{3}

Decimal

t = 8, t = 5.33333 \dots

Solution steps

3 t^{2} - 40 t + 128 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 3 \cdot 128}{2 \cdot 3}

(- 40)^{2} - 4 \cdot 3 \cdot 128 = 8

t_{1, 2} = \frac{- ( - 40 ) \pm 8}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- ( - 40 ) + 8}{2 \cdot 3}, t_{2} = \frac{- ( - 40 ) - 8}{2 \cdot 3}

t = \frac{- ( - 40 ) + 8}{2 \cdot 3} : 8

t = \frac{- ( - 40 ) - 8}{2 \cdot 3} : \frac{16}{3}

The solutions to the quadratic equation are:

t = 8, t = \frac{16}{3}

4 x^{2} + 36 x + 27 = 7 x^{2} + 54 x + 51

6 n^{2} = 2

x^{2} + 4 \times (- 21) = 0

4 y^{2} = - 4 y - 1

76 = 3 m^{2} + 10

What are the solutions to the equation 3t^2-40t+128=0 ?
The solutions to the equation 3t^2-40t+128=0 are t=8,t= 16/3
Find the zeros of 3t^2-40t+128=0
The zeros of 3t^2-40t+128=0 are t=8,t= 16/3