What are the solutions to the equation -25t^2+200t+400=0 ?

The solutions to the equation -25t^2+200t+400=0 are t=-4(sqrt(2)-1),t=4(1+sqrt(2))

Find the zeros of -25t^2+200t+400=0

The zeros of -25t^2+200t+400=0 are t=-4(sqrt(2)-1),t=4(1+sqrt(2))

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- 25 t^{2} + 200 t + 400 = 0

t = - 4 (2 - 1), t = 4 (1 + 2)

Decimal Notation

t = - 1.65685 \dots, t = 9.65685 \dots

Solution steps

- 25 t^{2} + 200 t + 400 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 200 \pm 20 0 ^{2} - 4 ( - 25 ) \cdot 400}{2 ( - 25 )}

20 0^{2} - 4 (- 25) \cdot 400 = 2002

t_{1, 2} = \frac{- 200 \pm 200 2}{2 ( - 25 )}

Separate the solutions

t_{1} = \frac{- 200 + 200 2}{2 ( - 25 )}, t_{2} = \frac{- 200 - 200 2}{2 ( - 25 )}

t = \frac{- 200 + 200 2}{2 ( - 25 )} : - 4 (2 - 1)

t = \frac{- 200 - 200 2}{2 ( - 25 )} : 4 (1 + 2)

The solutions to the quadratic equation are:

t = - 4 (2 - 1), t = 4 (1 + 2)

x^{2} = 5 - x

6 x - x^{2} = 0

x^{2} - 23 x + 126

\frac{3}{8} π \cdot 81 + \frac{27}{2} = 27 (\frac{9}{8} π + 1)

What are the solutions to the equation -25t^2+200t+400=0 ?
The solutions to the equation -25t^2+200t+400=0 are t=-4(sqrt(2)-1),t=4(1+sqrt(2))
Find the zeros of -25t^2+200t+400=0
The zeros of -25t^2+200t+400=0 are t=-4(sqrt(2)-1),t=4(1+sqrt(2))