What are the solutions to the equation -16t^2+80t+32=0 ?

The solutions to the equation -16t^2+80t+32=0 are t=-(-5+sqrt(33))/2 ,t=(5+sqrt(33))/2

Find the zeros of -16t^2+80t+32=0

The zeros of -16t^2+80t+32=0 are t=-(-5+sqrt(33))/2 ,t=(5+sqrt(33))/2

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- 16 t^{2} + 80 t + 32 = 0

t = - \frac{- 5 + 33}{2}, t = \frac{5 + 33}{2}

Solution steps

- 16 t^{2} + 80 t + 32 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 16 ) \cdot 32}{2 ( - 16 )}

8 0^{2} - 4 (- 16) \cdot 32 = 1633

t_{1, 2} = \frac{- 80 \pm 16 33}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 80 + 16 33}{2 ( - 16 )}, t_{2} = \frac{- 80 - 16 33}{2 ( - 16 )}

t = \frac{- 80 + 16 33}{2 ( - 16 )} : - \frac{- 5 + 33}{2}

t = \frac{- 80 - 16 33}{2 ( - 16 )} : \frac{5 + 33}{2}

The solutions to the quadratic equation are:

t = - \frac{- 5 + 33}{2}, t = \frac{5 + 33}{2}

\frac{6 n}{5} + \frac{1}{2} = \frac{9 n - 1}{10}

0. 5^{\frac{1}{3}}

\frac{a ^{2} b ^{- 7}}{a ^{- 4} b ^{- 2}}

(\frac{1}{16})^{\frac{1}{4}}

13 x + 7 = 5 x - 20

What are the solutions to the equation -16t^2+80t+32=0 ?
The solutions to the equation -16t^2+80t+32=0 are t=-(-5+sqrt(33))/2 ,t=(5+sqrt(33))/2
Find the zeros of -16t^2+80t+32=0
The zeros of -16t^2+80t+32=0 are t=-(-5+sqrt(33))/2 ,t=(5+sqrt(33))/2