What are the solutions to the equation 38=-16t^2+80t+102 ?

The solutions to the equation 38=-16t^2+80t+102 are t=-(-5+sqrt(41))/2 ,t=(5+sqrt(41))/2

Find the zeros of 38=-16t^2+80t+102

The zeros of 38=-16t^2+80t+102 are t=-(-5+sqrt(41))/2 ,t=(5+sqrt(41))/2

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38 = - 16 t^{2} + 80 t + 102

t = - \frac{- 5 + 41}{2}, t = \frac{5 + 41}{2}

Solution steps

38 = - 16 t^{2} + 80 t + 102

Switch sides

- 16 t^{2} + 80 t + 102 = 38

Move

38

to the left side

- 16 t^{2} + 80 t + 64 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 16 ) \cdot 64}{2 ( - 16 )}

8 0^{2} - 4 (- 16) \cdot 64 = 1641

t_{1, 2} = \frac{- 80 \pm 16 41}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 80 + 16 41}{2 ( - 16 )}, t_{2} = \frac{- 80 - 16 41}{2 ( - 16 )}

t = \frac{- 80 + 16 41}{2 ( - 16 )} : - \frac{- 5 + 41}{2}

t = \frac{- 80 - 16 41}{2 ( - 16 )} : \frac{5 + 41}{2}

The solutions to the quadratic equation are:

t = - \frac{- 5 + 41}{2}, t = \frac{5 + 41}{2}

- 9 \leq x + 5 \leq 15

16 c^{2} + 56 c + 49

(x + 4)^{2} + 5 (x + 4)

4^{(\frac{3}{2})}

15 v x^{3} y^{5} + 20 v^{6} x^{9}

What are the solutions to the equation 38=-16t^2+80t+102 ?
The solutions to the equation 38=-16t^2+80t+102 are t=-(-5+sqrt(41))/2 ,t=(5+sqrt(41))/2
Find the zeros of 38=-16t^2+80t+102
The zeros of 38=-16t^2+80t+102 are t=-(-5+sqrt(41))/2 ,t=(5+sqrt(41))/2