What are the solutions to the equation 180=-16t^2+120t ?

The solutions to the equation 180=-16t^2+120t are t=(3(5-sqrt(5)))/4 ,t=(3(5+sqrt(5)))/4

Find the zeros of 180=-16t^2+120t

The zeros of 180=-16t^2+120t are t=(3(5-sqrt(5)))/4 ,t=(3(5+sqrt(5)))/4

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180 = - 16 t^{2} + 120 t

t = \frac{3 ( 5 - 5 )}{4}, t = \frac{3 ( 5 + 5 )}{4}

Decimal Notation

t = 2.07294 \dots, t = 5.42705 \dots

Solution steps

180 = - 16 t^{2} + 120 t

Switch sides

- 16 t^{2} + 120 t = 180

Move

180

to the left side

- 16 t^{2} + 120 t - 180 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 16 ) ( - 180 )}{2 ( - 16 )}

12 0^{2} - 4 (- 16) (- 180) = 245

t_{1, 2} = \frac{- 120 \pm 24 5}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 120 + 24 5}{2 ( - 16 )}, t_{2} = \frac{- 120 - 24 5}{2 ( - 16 )}

t = \frac{- 120 + 24 5}{2 ( - 16 )} : \frac{3 ( 5 - 5 )}{4}

t = \frac{- 120 - 24 5}{2 ( - 16 )} : \frac{3 ( 5 + 5 )}{4}

The solutions to the quadratic equation are:

t = \frac{3 ( 5 - 5 )}{4}, t = \frac{3 ( 5 + 5 )}{4}

x, 3 x^{2} + 4 x - 1 = 0

7 x^{2} - 6 x + 5 = - 4

36 x^{2} = 9

18 x^{2} + 50 = 0

x^{2} = 23 + 2 x

What are the solutions to the equation 180=-16t^2+120t ?
The solutions to the equation 180=-16t^2+120t are t=(3(5-sqrt(5)))/4 ,t=(3(5+sqrt(5)))/4
Find the zeros of 180=-16t^2+120t
The zeros of 180=-16t^2+120t are t=(3(5-sqrt(5)))/4 ,t=(3(5+sqrt(5)))/4