What are the solutions to the equation 16t^2-64t+48=0 ?

The solutions to the equation 16t^2-64t+48=0 are t=3,t=1

Find the zeros of 16t^2-64t+48=0

The zeros of 16t^2-64t+48=0 are t=3,t=1

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16 t^{2} - 64 t + 48 = 0

t = 3, t = 1

Solution steps

16 t^{2} - 64 t + 48 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 64 ) \pm ( - 64 ) ^{2} - 4 \cdot 16 \cdot 48}{2 \cdot 16}

(- 64)^{2} - 4 \cdot 16 \cdot 48 = 32

t_{1, 2} = \frac{- ( - 64 ) \pm 32}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- ( - 64 ) + 32}{2 \cdot 16}, t_{2} = \frac{- ( - 64 ) - 32}{2 \cdot 16}

t = \frac{- ( - 64 ) + 32}{2 \cdot 16} : 3

t = \frac{- ( - 64 ) - 32}{2 \cdot 16} : 1

The solutions to the quadratic equation are:

t = 3, t = 1

2 x^{2} - 3 x = - 6

a, u^{3} = a^{2} t

x^{2} + 20 x = 400

2 x^{2} - 8 x - 21 = 0

4 x^{2} - 371 = 29

What are the solutions to the equation 16t^2-64t+48=0 ?
The solutions to the equation 16t^2-64t+48=0 are t=3,t=1
Find the zeros of 16t^2-64t+48=0
The zeros of 16t^2-64t+48=0 are t=3,t=1