What are the solutions to the equation 150=200-20t+2t^2 ?

The solutions to the equation 150=200-20t+2t^2 are t=5

Find the zeros of 150=200-20t+2t^2

The zeros of 150=200-20t+2t^2 are t=5

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150 = 200 - 20 t + 2 t^{2}

t = 5

Solution steps

150 = 200 - 20 t + 2 t^{2}

Switch sides

200 - 20 t + 2 t^{2} = 150

Move

150

to the left side

2 t^{2} - 20 t + 50 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 \cdot 50}{2 \cdot 2}

(- 20)^{2} - 4 \cdot 2 \cdot 50 = 0

t_{1, 2} = \frac{- ( - 20 ) \pm 0}{2 \cdot 2}

t = \frac{- ( - 20 )}{2 \cdot 2}

\frac{- ( - 20 )}{2 \cdot 2} = 5

t = 5

The solution to the quadratic equation is:

t = 5

m^{2} - 6 m = 7

x^{2} + 8 x - 28 = 0

2 x^{2} + 12 x = 6

5 x^{2} = 24 - x^{2}

m, m^{2} - 15 m + 56 = 0

What are the solutions to the equation 150=200-20t+2t^2 ?
The solutions to the equation 150=200-20t+2t^2 are t=5
Find the zeros of 150=200-20t+2t^2
The zeros of 150=200-20t+2t^2 are t=5