What are the solutions to the equation 0=-16t^2+80t+3 ?

The solutions to the equation 0=-16t^2+80t+3 are t=-(-10+sqrt(103))/4 ,t=(10+sqrt(103))/4

Find the zeros of 0=-16t^2+80t+3

The zeros of 0=-16t^2+80t+3 are t=-(-10+sqrt(103))/4 ,t=(10+sqrt(103))/4

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0 = - 16 t^{2} + 80 t + 3

t = - \frac{- 10 + 103}{4}, t = \frac{10 + 103}{4}

Decimal Notation

t = - 0.03722 \dots, t = 5.03722 \dots

Solution steps

0 = - 16 t^{2} + 80 t + 3

Switch sides

- 16 t^{2} + 80 t + 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

8 0^{2} - 4 (- 16) \cdot 3 = 8103

t_{1, 2} = \frac{- 80 \pm 8 103}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 80 + 8 103}{2 ( - 16 )}, t_{2} = \frac{- 80 - 8 103}{2 ( - 16 )}

t = \frac{- 80 + 8 103}{2 ( - 16 )} : - \frac{- 10 + 103}{4}

t = \frac{- 80 - 8 103}{2 ( - 16 )} : \frac{10 + 103}{4}

The solutions to the quadratic equation are:

t = - \frac{- 10 + 103}{4}, t = \frac{10 + 103}{4}

1 5^{2} + 2 0^{2} = x^{2}

x (x + 1) = 182

∣ 3 x - 1 ∣ > 11

5 x + 4 \geq 7 x - 1

- x^{2} + x + 12 = 0

What are the solutions to the equation 0=-16t^2+80t+3 ?
The solutions to the equation 0=-16t^2+80t+3 are t=-(-10+sqrt(103))/4 ,t=(10+sqrt(103))/4
Find the zeros of 0=-16t^2+80t+3
The zeros of 0=-16t^2+80t+3 are t=-(-10+sqrt(103))/4 ,t=(10+sqrt(103))/4