What are the solutions to the equation 4t-0.8t^2=2.5 ?

The solutions to the equation 4t-0.8t^2=2.5 are t=(5(2-sqrt(2)))/4 ,t=(5(2+sqrt(2)))/4

Find the zeros of 4t-0.8t^2=2.5

The zeros of 4t-0.8t^2=2.5 are t=(5(2-sqrt(2)))/4 ,t=(5(2+sqrt(2)))/4

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4 t - 0.8 t^{2} = 2.5

t = \frac{5 ( 2 - 2 )}{4}, t = \frac{5 ( 2 + 2 )}{4}

Decimal Notation

t = 0.73223 \dots, t = 4.26776 \dots

Solution steps

4 t - 0.8 t^{2} = 2.5

Multiply both sides by

10

40 t - 8 t^{2} = 25

Move

25

to the left side

40 t - 8 t^{2} - 25 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 8 t^{2} + 40 t - 25 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 8 ) ( - 25 )}{2 ( - 8 )}

4 0^{2} - 4 (- 8) (- 25) = 202

t_{1, 2} = \frac{- 40 \pm 20 2}{2 ( - 8 )}

Separate the solutions

t_{1} = \frac{- 40 + 20 2}{2 ( - 8 )}, t_{2} = \frac{- 40 - 20 2}{2 ( - 8 )}

t = \frac{- 40 + 20 2}{2 ( - 8 )} : \frac{5 ( 2 - 2 )}{4}

t = \frac{- 40 - 20 2}{2 ( - 8 )} : \frac{5 ( 2 + 2 )}{4}

The solutions to the quadratic equation are:

t = \frac{5 ( 2 - 2 )}{4}, t = \frac{5 ( 2 + 2 )}{4}

- 18 + 5 x = - 12 x - 1

4 x^{2} + 3 x + 7 = 0

3 f + e = 6

\frac{( \frac{( 400 - 25 x ^{2} )}{4} )}{64} - \frac{( x ^{2} )}{25} = 1

3 - 12 \times 2 + 19

What are the solutions to the equation 4t-0.8t^2=2.5 ?
The solutions to the equation 4t-0.8t^2=2.5 are t=(5(2-sqrt(2)))/4 ,t=(5(2+sqrt(2)))/4
Find the zeros of 4t-0.8t^2=2.5
The zeros of 4t-0.8t^2=2.5 are t=(5(2-sqrt(2)))/4 ,t=(5(2+sqrt(2)))/4