What are the solutions to the equation 0=16t^2-140t+68 ?

The solutions to the equation 0=16t^2-140t+68 are t=(35+sqrt(953))/8 ,t=(35-sqrt(953))/8

Find the zeros of 0=16t^2-140t+68

The zeros of 0=16t^2-140t+68 are t=(35+sqrt(953))/8 ,t=(35-sqrt(953))/8

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0 = 16 t^{2} - 140 t + 68

t = \frac{35 + 953}{8}, t = \frac{35 - 953}{8}

Decimal Notation

t = 8.23383 \dots, t = 0.51616 \dots

Solution steps

0 = 16 t^{2} - 140 t + 68

Switch sides

16 t^{2} - 140 t + 68 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 140 ) \pm ( - 140 ) ^{2} - 4 \cdot 16 \cdot 68}{2 \cdot 16}

(- 140)^{2} - 4 \cdot 16 \cdot 68 = 4953

t_{1, 2} = \frac{- ( - 140 ) \pm 4 953}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- ( - 140 ) + 4 953}{2 \cdot 16}, t_{2} = \frac{- ( - 140 ) - 4 953}{2 \cdot 16}

t = \frac{- ( - 140 ) + 4 953}{2 \cdot 16} : \frac{35 + 953}{8}

t = \frac{- ( - 140 ) - 4 953}{2 \cdot 16} : \frac{35 - 953}{8}

The solutions to the quadratic equation are:

t = \frac{35 + 953}{8}, t = \frac{35 - 953}{8}

\frac{1}{2} = \frac{4}{5} x - \frac{1}{4}

\frac{2}{x} - 3 \leq 5

(x - 4)^{2} + 25 = 0

t^{2} + 7 t + 12 = 0

- 1 + \frac{1}{2}

What are the solutions to the equation 0=16t^2-140t+68 ?
The solutions to the equation 0=16t^2-140t+68 are t=(35+sqrt(953))/8 ,t=(35-sqrt(953))/8
Find the zeros of 0=16t^2-140t+68
The zeros of 0=16t^2-140t+68 are t=(35+sqrt(953))/8 ,t=(35-sqrt(953))/8