What are the solutions to the equation 16t^2-208t+576=0 ?

The solutions to the equation 16t^2-208t+576=0 are t=9,t=4

Find the zeros of 16t^2-208t+576=0

The zeros of 16t^2-208t+576=0 are t=9,t=4

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16 t^{2} - 208 t + 576 = 0

t = 9, t = 4

Solution steps

16 t^{2} - 208 t + 576 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 208 ) \pm ( - 208 ) ^{2} - 4 \cdot 16 \cdot 576}{2 \cdot 16}

(- 208)^{2} - 4 \cdot 16 \cdot 576 = 80

t_{1, 2} = \frac{- ( - 208 ) \pm 80}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- ( - 208 ) + 80}{2 \cdot 16}, t_{2} = \frac{- ( - 208 ) - 80}{2 \cdot 16}

t = \frac{- ( - 208 ) + 80}{2 \cdot 16} : 9

t = \frac{- ( - 208 ) - 80}{2 \cdot 16} : 4

The solutions to the quadratic equation are:

t = 9, t = 4

4 x^{2} - (a + 3) x + a = 0

x, 0 = x^{2} + 2 x

(x + 9)^{2} + 27 = - 5

64 k^{2} - 4 k = 0

x^{2} + 57 = 16 x

What are the solutions to the equation 16t^2-208t+576=0 ?
The solutions to the equation 16t^2-208t+576=0 are t=9,t=4
Find the zeros of 16t^2-208t+576=0
The zeros of 16t^2-208t+576=0 are t=9,t=4