What are the solutions to the equation 24t-0.8t^2=90 ?

The solutions to the equation 24t-0.8t^2=90 are t=(15(2-sqrt(2)))/2 ,t=(15(2+sqrt(2)))/2

Find the zeros of 24t-0.8t^2=90

The zeros of 24t-0.8t^2=90 are t=(15(2-sqrt(2)))/2 ,t=(15(2+sqrt(2)))/2

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24 t - 0.8 t^{2} = 90

t = \frac{15 ( 2 - 2 )}{2}, t = \frac{15 ( 2 + 2 )}{2}

Decimal Notation

t = 4.39339 \dots, t = 25.60660 \dots

Solution steps

24 t - 0.8 t^{2} = 90

Multiply both sides by

10

240 t - 8 t^{2} = 900

Move

900

to the left side

240 t - 8 t^{2} - 900 = 0

Write in the standard form

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

- 8 t^{2} + 240 t - 900 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 240 \pm 24 0 ^{2} - 4 ( - 8 ) ( - 900 )}{2 ( - 8 )}

24 0^{2} - 4 (- 8) (- 900) = 1202

t_{1, 2} = \frac{- 240 \pm 120 2}{2 ( - 8 )}

Separate the solutions

t_{1} = \frac{- 240 + 120 2}{2 ( - 8 )}, t_{2} = \frac{- 240 - 120 2}{2 ( - 8 )}

t = \frac{- 240 + 120 2}{2 ( - 8 )} : \frac{15 ( 2 - 2 )}{2}

t = \frac{- 240 - 120 2}{2 ( - 8 )} : \frac{15 ( 2 + 2 )}{2}

The solutions to the quadratic equation are:

t = \frac{15 ( 2 - 2 )}{2}, t = \frac{15 ( 2 + 2 )}{2}

32 - 162 k^{2} = 0

11 c - c d

2 u^{2} - 64 = 0

\frac{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}{3 \cdot 3 \cdot 3 \cdot 3}

2 k^{2} - 8 k - 90

What are the solutions to the equation 24t-0.8t^2=90 ?
The solutions to the equation 24t-0.8t^2=90 are t=(15(2-sqrt(2)))/2 ,t=(15(2+sqrt(2)))/2
Find the zeros of 24t-0.8t^2=90
The zeros of 24t-0.8t^2=90 are t=(15(2-sqrt(2)))/2 ,t=(15(2+sqrt(2)))/2