What are the solutions to the equation -16t^2+80t=32 ?

The solutions to the equation -16t^2+80t=32 are t=(5-sqrt(17))/2 ,t=(5+sqrt(17))/2

Find the zeros of -16t^2+80t=32

The zeros of -16t^2+80t=32 are t=(5-sqrt(17))/2 ,t=(5+sqrt(17))/2

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- 16 t^{2} + 80 t = 32

t = \frac{5 - 17}{2}, t = \frac{5 + 17}{2}

Decimal Notation

t = 0.43844 \dots, t = 4.56155 \dots

Solution steps

- 16 t^{2} + 80 t = 32

Move

32

to the left side

- 16 t^{2} + 80 t - 32 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 16 ) ( - 32 )}{2 ( - 16 )}

8 0^{2} - 4 (- 16) (- 32) = 1617

t_{1, 2} = \frac{- 80 \pm 16 17}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 80 + 16 17}{2 ( - 16 )}, t_{2} = \frac{- 80 - 16 17}{2 ( - 16 )}

t = \frac{- 80 + 16 17}{2 ( - 16 )} : \frac{5 - 17}{2}

t = \frac{- 80 - 16 17}{2 ( - 16 )} : \frac{5 + 17}{2}

The solutions to the quadratic equation are:

t = \frac{5 - 17}{2}, t = \frac{5 + 17}{2}

∣ w ∣ \geq 5

(- 6 a^{2} b) \cdot (\frac{1}{2} ab)

3 x^{2} + 2 x - 456

x^{2} + 2 x^{1} + 1 = 0

x^{2} + 8 x = 9

What are the solutions to the equation -16t^2+80t=32 ?
The solutions to the equation -16t^2+80t=32 are t=(5-sqrt(17))/2 ,t=(5+sqrt(17))/2
Find the zeros of -16t^2+80t=32
The zeros of -16t^2+80t=32 are t=(5-sqrt(17))/2 ,t=(5+sqrt(17))/2