What are the solutions to the equation 200= 2/125 t^2-16/5 t+320 ?

The solutions to the equation 200= 2/125 t^2-16/5 t+320 are t=150,t=50

Find the zeros of 200= 2/125 t^2-16/5 t+320

The zeros of 200= 2/125 t^2-16/5 t+320 are t=150,t=50

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200 = \frac{2}{125} t^{2} - \frac{16}{5} t + 320

t = 150, t = 50

Solution steps

200 = \frac{2}{125} t^{2} - \frac{16}{5} t + 320

Find Least Common Multiplier of

125, 5 : 125

Multiply by LCM=

125

200 \cdot 125 = \frac{2}{125} t^{2} \cdot 125 - \frac{16}{5} t \cdot 125 + 320 \cdot 125

Simplify

25000 = 2 t^{2} - 400 t + 40000

Switch sides

2 t^{2} - 400 t + 40000 = 25000

Move

25000

to the left side

2 t^{2} - 400 t + 15000 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 400 ) \pm ( - 400 ) ^{2} - 4 \cdot 2 \cdot 15000}{2 \cdot 2}

(- 400)^{2} - 4 \cdot 2 \cdot 15000 = 200

t_{1, 2} = \frac{- ( - 400 ) \pm 200}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- ( - 400 ) + 200}{2 \cdot 2}, t_{2} = \frac{- ( - 400 ) - 200}{2 \cdot 2}

t = \frac{- ( - 400 ) + 200}{2 \cdot 2} : 150

t = \frac{- ( - 400 ) - 200}{2 \cdot 2} : 50

The solutions to the quadratic equation are:

t = 150, t = 50

6 x^{2} - x^{2} + 2 x = x

P^{2} = 2.97370166 \cdot 1 0^{- 19} \cdot 149597870700 0^{3}

x, (f \circ g) (x) = 3 x^{2} + 1

1 5^{2} + 1 4^{2} = x^{2}

x^{2} - 4 x = - x

What are the solutions to the equation 200= 2/125 t^2-16/5 t+320 ?
The solutions to the equation 200= 2/125 t^2-16/5 t+320 are t=150,t=50
Find the zeros of 200= 2/125 t^2-16/5 t+320
The zeros of 200= 2/125 t^2-16/5 t+320 are t=150,t=50