What are the solutions to the equation -16t^2+68t-27=0 ?

The solutions to the equation -16t^2+68t-27=0 are t=(17-sqrt(181))/8 ,t=(17+sqrt(181))/8

Find the zeros of -16t^2+68t-27=0

The zeros of -16t^2+68t-27=0 are t=(17-sqrt(181))/8 ,t=(17+sqrt(181))/8

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- 16 t^{2} + 68 t - 27 = 0

t = \frac{17 - 181}{8}, t = \frac{17 + 181}{8}

Decimal Notation

t = 0.44329 \dots, t = 3.80670 \dots

Solution steps

- 16 t^{2} + 68 t - 27 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 68 \pm 6 8 ^{2} - 4 ( - 16 ) ( - 27 )}{2 ( - 16 )}

6 8^{2} - 4 (- 16) (- 27) = 4181

t_{1, 2} = \frac{- 68 \pm 4 181}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 68 + 4 181}{2 ( - 16 )}, t_{2} = \frac{- 68 - 4 181}{2 ( - 16 )}

t = \frac{- 68 + 4 181}{2 ( - 16 )} : \frac{17 - 181}{8}

t = \frac{- 68 - 4 181}{2 ( - 16 )} : \frac{17 + 181}{8}

The solutions to the quadratic equation are:

t = \frac{17 - 181}{8}, t = \frac{17 + 181}{8}

∣ x - 1 ∣ < 2 x

(2 x - 3) (2 x + 3) - 135 = 0

2 x^{2} + 3 x - 5 = (x + 1)^{2}

(x - 1)^{2} > 0

- 1 j + 1 > 10

What are the solutions to the equation -16t^2+68t-27=0 ?
The solutions to the equation -16t^2+68t-27=0 are t=(17-sqrt(181))/8 ,t=(17+sqrt(181))/8
Find the zeros of -16t^2+68t-27=0
The zeros of -16t^2+68t-27=0 are t=(17-sqrt(181))/8 ,t=(17+sqrt(181))/8