What are the solutions to the equation 3+40t-16t^2=16 ?

The solutions to the equation 3+40t-16t^2=16 are t=(5-2sqrt(3))/4 ,t=(5+2sqrt(3))/4

Find the zeros of 3+40t-16t^2=16

The zeros of 3+40t-16t^2=16 are t=(5-2sqrt(3))/4 ,t=(5+2sqrt(3))/4

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3 + 40 t - 16 t^{2} = 16

t = \frac{5 - 2 3}{4}, t = \frac{5 + 2 3}{4}

Decimal Notation

t = 0.38397 \dots, t = 2.11602 \dots

Solution steps

3 + 40 t - 16 t^{2} = 16

Move

16

to the left side

- 16 t^{2} + 40 t - 13 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) ( - 13 )}{2 ( - 16 )}

4 0^{2} - 4 (- 16) (- 13) = 163

t_{1, 2} = \frac{- 40 \pm 16 3}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 40 + 16 3}{2 ( - 16 )}, t_{2} = \frac{- 40 - 16 3}{2 ( - 16 )}

t = \frac{- 40 + 16 3}{2 ( - 16 )} : \frac{5 - 2 3}{4}

t = \frac{- 40 - 16 3}{2 ( - 16 )} : \frac{5 + 2 3}{4}

The solutions to the quadratic equation are:

t = \frac{5 - 2 3}{4}, t = \frac{5 + 2 3}{4}

(x^{\frac{1}{3}})^{4}

∣ x - 5 ∣ < ∣ x + 1 ∣

\frac{1}{3 2 ^{\frac{1}{5}}}

v^{2} + 14 v + 19 = 0

\frac{8 x ^{3} y ^{5}}{2 x y ^{2}}

What are the solutions to the equation 3+40t-16t^2=16 ?
The solutions to the equation 3+40t-16t^2=16 are t=(5-2sqrt(3))/4 ,t=(5+2sqrt(3))/4
Find the zeros of 3+40t-16t^2=16
The zeros of 3+40t-16t^2=16 are t=(5-2sqrt(3))/4 ,t=(5+2sqrt(3))/4