What are the solutions to the equation 16t^2-128t+240=0 ?

The solutions to the equation 16t^2-128t+240=0 are t=5,t=3

Find the zeros of 16t^2-128t+240=0

The zeros of 16t^2-128t+240=0 are t=5,t=3

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16 t^{2} - 128 t + 240 = 0

t = 5, t = 3

Solution steps

16 t^{2} - 128 t + 240 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 128 ) \pm ( - 128 ) ^{2} - 4 \cdot 16 \cdot 240}{2 \cdot 16}

(- 128)^{2} - 4 \cdot 16 \cdot 240 = 32

t_{1, 2} = \frac{- ( - 128 ) \pm 32}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- ( - 128 ) + 32}{2 \cdot 16}, t_{2} = \frac{- ( - 128 ) - 32}{2 \cdot 16}

t = \frac{- ( - 128 ) + 32}{2 \cdot 16} : 5

t = \frac{- ( - 128 ) - 32}{2 \cdot 16} : 3

The solutions to the quadratic equation are:

t = 5, t = 3

18 + 6 y - y^{2} = 0

4 x^{2} + 4 x + 7 = 0

0 = - x^{2} + 6 x

0.4 x^{2} - 1.2 x - 0.36 = 0

(6 t + 1)^{2} + 3 = 0

What are the solutions to the equation 16t^2-128t+240=0 ?
The solutions to the equation 16t^2-128t+240=0 are t=5,t=3
Find the zeros of 16t^2-128t+240=0
The zeros of 16t^2-128t+240=0 are t=5,t=3