What are the solutions to the equation 0=40t-16t^2+80 ?

The solutions to the equation 0=40t-16t^2+80 are t=-(-5+sqrt(105))/4 ,t=(5+sqrt(105))/4

Find the zeros of 0=40t-16t^2+80

The zeros of 0=40t-16t^2+80 are t=-(-5+sqrt(105))/4 ,t=(5+sqrt(105))/4

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0 = 40 t - 16 t^{2} + 80

t = - \frac{- 5 + 105}{4}, t = \frac{5 + 105}{4}

Decimal Notation

t = - 1.31173 \dots, t = 3.81173 \dots

Solution steps

0 = 40 t - 16 t^{2} + 80

Switch sides

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

Write in the standard form

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

- 16 t^{2} + 40 t + 80 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) \cdot 80}{2 ( - 16 )}

4 0^{2} - 4 (- 16) \cdot 80 = 8105

t_{1, 2} = \frac{- 40 \pm 8 105}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 40 + 8 105}{2 ( - 16 )}, t_{2} = \frac{- 40 - 8 105}{2 ( - 16 )}

t = \frac{- 40 + 8 105}{2 ( - 16 )} : - \frac{- 5 + 105}{4}

t = \frac{- 40 - 8 105}{2 ( - 16 )} : \frac{5 + 105}{4}

The solutions to the quadratic equation are:

t = - \frac{- 5 + 105}{4}, t = \frac{5 + 105}{4}

(y^{- 5})^{- 10} y^{10}

- 3 x^{2} + 10 x - 3 = 0

x + 17 \leq - 3

(x - 1)^{2} + (1 - x) \times (x + 2) = 0

x^{2} - 7 x = 8

What are the solutions to the equation 0=40t-16t^2+80 ?
The solutions to the equation 0=40t-16t^2+80 are t=-(-5+sqrt(105))/4 ,t=(5+sqrt(105))/4
Find the zeros of 0=40t-16t^2+80
The zeros of 0=40t-16t^2+80 are t=-(-5+sqrt(105))/4 ,t=(5+sqrt(105))/4