What are the solutions to the equation -16t^2+40t+3=0 ?

The solutions to the equation -16t^2+40t+3=0 are t=-(-5+2sqrt(7))/4 ,t=(5+2sqrt(7))/4

Find the zeros of -16t^2+40t+3=0

The zeros of -16t^2+40t+3=0 are t=-(-5+2sqrt(7))/4 ,t=(5+2sqrt(7))/4

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- 16 t^{2} + 40 t + 3 = 0

t = - \frac{- 5 + 2 7}{4}, t = \frac{5 + 2 7}{4}

Decimal Notation

t = - 0.07287 \dots, t = 2.57287 \dots

Solution steps

- 16 t^{2} + 40 t + 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

4 0^{2} - 4 (- 16) \cdot 3 = 167

t_{1, 2} = \frac{- 40 \pm 16 7}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 40 + 16 7}{2 ( - 16 )}, t_{2} = \frac{- 40 - 16 7}{2 ( - 16 )}

t = \frac{- 40 + 16 7}{2 ( - 16 )} : - \frac{- 5 + 2 7}{4}

t = \frac{- 40 - 16 7}{2 ( - 16 )} : \frac{5 + 2 7}{4}

The solutions to the quadratic equation are:

t = - \frac{- 5 + 2 7}{4}, t = \frac{5 + 2 7}{4}

x^{2} - x + 20 = 0

i^{\frac{3}{2}}

∣ x - 5 ∣ < 5

- 3 (w + 3) < 9 - 3 w

mn + n

What are the solutions to the equation -16t^2+40t+3=0 ?
The solutions to the equation -16t^2+40t+3=0 are t=-(-5+2sqrt(7))/4 ,t=(5+2sqrt(7))/4
Find the zeros of -16t^2+40t+3=0
The zeros of -16t^2+40t+3=0 are t=-(-5+2sqrt(7))/4 ,t=(5+2sqrt(7))/4