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

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

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

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

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180 = 16 t^{2} + 120 t

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

Decimal Notation

t = 1.28115 \dots, t = - 8.78115 \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 \cdot 16 ( - 180 )}{2 \cdot 16}

12 0^{2} - 4 \cdot 16 (- 180) = 725

t_{1, 2} = \frac{- 120 \pm 72 5}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- 120 + 72 5}{2 \cdot 16}, t_{2} = \frac{- 120 - 72 5}{2 \cdot 16}

t = \frac{- 120 + 72 5}{2 \cdot 16} : \frac{3 ( 3 5 - 5 )}{4}

t = \frac{- 120 - 72 5}{2 \cdot 16} : - \frac{3 ( 5 + 3 5 )}{4}

The solutions to the quadratic equation are:

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

5 x^{2} + 75 x = 0

0 = 4 x^{2} + 4 x - 14

2 x^{2} + 20 x = - 3

2 x^{2} + 12 x + 3 = 15

12 x^{2} - 6 x + 10 = 0

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