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

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

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

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

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

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

Solution steps

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

Multiply both sides by

10

- 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}

1010 x^{2} + 1009 x - 1008 = 0

- 7 = 2 x^{2} - 5

- 2 x^{2} - 5 x + 27 = 0

0.2 x^{2} - 0.4 x - 15 = 0

x = 2 x^{2} + 5

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