What are the solutions to the equation 0=-16t^2+80t+4 ?

The solutions to the equation 0=-16t^2+80t+4 are t=-(-5+sqrt(26))/2 ,t=(5+sqrt(26))/2

Find the zeros of 0=-16t^2+80t+4

The zeros of 0=-16t^2+80t+4 are t=-(-5+sqrt(26))/2 ,t=(5+sqrt(26))/2

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0 = - 16 t^{2} + 80 t + 4

t = - \frac{- 5 + 26}{2}, t = \frac{5 + 26}{2}

Decimal Notation

t = - 0.04950 \dots, t = 5.04950 \dots

Solution steps

0 = - 16 t^{2} + 80 t + 4

Switch sides

- 16 t^{2} + 80 t + 4 = 0

Solve with the quadratic formula

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

8 0^{2} - 4 (- 16) \cdot 4 = 1626

t_{1, 2} = \frac{- 80 \pm 16 26}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 80 + 16 26}{2 ( - 16 )}, t_{2} = \frac{- 80 - 16 26}{2 ( - 16 )}

t = \frac{- 80 + 16 26}{2 ( - 16 )} : - \frac{- 5 + 26}{2}

t = \frac{- 80 - 16 26}{2 ( - 16 )} : \frac{5 + 26}{2}

The solutions to the quadratic equation are:

t = - \frac{- 5 + 26}{2}, t = \frac{5 + 26}{2}

x, x^{2} y^{2} + 1 = 2 x y

29 (x + 1) + 3 = - x^{2} + 27 x + 6

\frac{350 x ^{2}}{2} = 6.64 \cdot 1257 \cdot (540 - x)

2 (3 h - 1) = 4 - 2 h (h + 5)

(2 x - 3)^{2} = (x + 1) (2 x - 3)

What are the solutions to the equation 0=-16t^2+80t+4 ?
The solutions to the equation 0=-16t^2+80t+4 are t=-(-5+sqrt(26))/2 ,t=(5+sqrt(26))/2
Find the zeros of 0=-16t^2+80t+4
The zeros of 0=-16t^2+80t+4 are t=-(-5+sqrt(26))/2 ,t=(5+sqrt(26))/2