What are the solutions to the equation -0.9t^2+18t=0 ?

The solutions to the equation -0.9t^2+18t=0 are t=0,t=20

Find the zeros of -0.9t^2+18t=0

The zeros of -0.9t^2+18t=0 are t=0,t=20

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- 0.9 t^{2} + 18 t = 0

t = 0, t = 20

Solution steps

- 0.9 t^{2} + 18 t = 0

Multiply both sides by

10

- 9 t^{2} + 180 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 180 \pm 18 0 ^{2} - 4 ( - 9 ) \cdot 0}{2 ( - 9 )}

18 0^{2} - 4 (- 9) \cdot 0 = 180

t_{1, 2} = \frac{- 180 \pm 180}{2 ( - 9 )}

Separate the solutions

t_{1} = \frac{- 180 + 180}{2 ( - 9 )}, t_{2} = \frac{- 180 - 180}{2 ( - 9 )}

t = \frac{- 180 + 180}{2 ( - 9 )} : 0

t = \frac{- 180 - 180}{2 ( - 9 )} : 20

The solutions to the quadratic equation are:

t = 0, t = 20

e^{x + i y} + e^{- x - i y}

3 x^{2} + 7 x \leq 6

r^{2} - 1 r - 20 = 0

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

- x^{2} + 23 x - 132 = 0

What are the solutions to the equation -0.9t^2+18t=0 ?
The solutions to the equation -0.9t^2+18t=0 are t=0,t=20
Find the zeros of -0.9t^2+18t=0
The zeros of -0.9t^2+18t=0 are t=0,t=20