What are the solutions to the equation 24=-0.5t^2+3.2t+22 ?

The solutions to the equation 24=-0.5t^2+3.2t+22 are t=(2(8-sqrt(39)))/5 ,t=(2(8+sqrt(39)))/5

Find the zeros of 24=-0.5t^2+3.2t+22

The zeros of 24=-0.5t^2+3.2t+22 are t=(2(8-sqrt(39)))/5 ,t=(2(8+sqrt(39)))/5

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24 = - 0.5 t^{2} + 3.2 t + 22

t = \frac{2 ( 8 - 39 )}{5}, t = \frac{2 ( 8 + 39 )}{5}

Decimal Notation

t = 0.70200 \dots, t = 5.69799 \dots

Solution steps

24 = - 0.5 t^{2} + 3.2 t + 22

Multiply both sides by

10

240 = - 5 t^{2} + 32 t + 220

Switch sides

- 5 t^{2} + 32 t + 220 = 240

Move

240

to the left side

- 5 t^{2} + 32 t - 20 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 5 ) ( - 20 )}{2 ( - 5 )}

3 2^{2} - 4 (- 5) (- 20) = 439

t_{1, 2} = \frac{- 32 \pm 4 39}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 32 + 4 39}{2 ( - 5 )}, t_{2} = \frac{- 32 - 4 39}{2 ( - 5 )}

t = \frac{- 32 + 4 39}{2 ( - 5 )} : \frac{2 ( 8 - 39 )}{5}

t = \frac{- 32 - 4 39}{2 ( - 5 )} : \frac{2 ( 8 + 39 )}{5}

The solutions to the quadratic equation are:

t = \frac{2 ( 8 - 39 )}{5}, t = \frac{2 ( 8 + 39 )}{5}

b^{2} = 34

50 = \frac{3 x ( 20 - x )}{2}

4 x^{2} + 36 x + 31

15 x^{2} - 6 = - 13 x

x, x^{2} + 9 b^{2} = 0

What are the solutions to the equation 24=-0.5t^2+3.2t+22 ?
The solutions to the equation 24=-0.5t^2+3.2t+22 are t=(2(8-sqrt(39)))/5 ,t=(2(8+sqrt(39)))/5
Find the zeros of 24=-0.5t^2+3.2t+22
The zeros of 24=-0.5t^2+3.2t+22 are t=(2(8-sqrt(39)))/5 ,t=(2(8+sqrt(39)))/5