What are the solutions to the equation -16t^2+40=40(t) ?

The solutions to the equation -16t^2+40=40(t) are t=-(5+sqrt(65))/4 ,t=(sqrt(65)-5)/4

Find the zeros of -16t^2+40=40(t)

The zeros of -16t^2+40=40(t) are t=-(5+sqrt(65))/4 ,t=(sqrt(65)-5)/4

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- 16 t^{2} + 40 = 40 (t)

t = - \frac{5 + 65}{4}, t = \frac{65 - 5}{4}

Decimal Notation

t = - 3.26556 \dots, t = 0.76556 \dots

Solution steps

- 16 t^{2} + 40 = 40 (t)

Refine

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

Move

40 t

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 40)^{2} - 4 (- 16) \cdot 40 = 865

t_{1, 2} = \frac{- ( - 40 ) \pm 8 65}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- ( - 40 ) + 8 65}{2 ( - 16 )}, t_{2} = \frac{- ( - 40 ) - 8 65}{2 ( - 16 )}

t = \frac{- ( - 40 ) + 8 65}{2 ( - 16 )} : - \frac{5 + 65}{4}

t = \frac{- ( - 40 ) - 8 65}{2 ( - 16 )} : \frac{65 - 5}{4}

The solutions to the quadratic equation are:

t = - \frac{5 + 65}{4}, t = \frac{65 - 5}{4}

(x + 5)^{2} - 18 = 63

x, 52 x^{2} + 72 x y + 73 y^{2} - 40 x + 30 y = 75

2 z^{2} - 3 z + 3 = 0

r^{2} + 3 r + r = 0

x^{2} + 5 x = 0

What are the solutions to the equation -16t^2+40=40(t) ?
The solutions to the equation -16t^2+40=40(t) are t=-(5+sqrt(65))/4 ,t=(sqrt(65)-5)/4
Find the zeros of -16t^2+40=40(t)
The zeros of -16t^2+40=40(t) are t=-(5+sqrt(65))/4 ,t=(sqrt(65)-5)/4