What are the solutions to the equation -16t^2+40t+3=10 ?

The solutions to the equation -16t^2+40t+3=10 are t=(5-3sqrt(2))/4 ,t=(5+3sqrt(2))/4

Find the zeros of -16t^2+40t+3=10

The zeros of -16t^2+40t+3=10 are t=(5-3sqrt(2))/4 ,t=(5+3sqrt(2))/4

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- 16 t^{2} + 40 t + 3 = 10

t = \frac{5 - 3 2}{4}, t = \frac{5 + 3 2}{4}

Decimal Notation

t = 0.18933 \dots, t = 2.31066 \dots

Solution steps

- 16 t^{2} + 40 t + 3 = 10

Move

10

to the left side

- 16 t^{2} + 40 t - 7 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) ( - 7 )}{2 ( - 16 )}

4 0^{2} - 4 (- 16) (- 7) = 242

t_{1, 2} = \frac{- 40 \pm 24 2}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 40 + 24 2}{2 ( - 16 )}, t_{2} = \frac{- 40 - 24 2}{2 ( - 16 )}

t = \frac{- 40 + 24 2}{2 ( - 16 )} : \frac{5 - 3 2}{4}

t = \frac{- 40 - 24 2}{2 ( - 16 )} : \frac{5 + 3 2}{4}

The solutions to the quadratic equation are:

t = \frac{5 - 3 2}{4}, t = \frac{5 + 3 2}{4}

3 m^{2} - 6 mn + 4 m - 8 n

\frac{b ^{8}}{b ^{2}}

3 (10 - x) = 21

(1 - x)^{2} < \frac{9}{25}

49 x^{4} y^{2} - 64 w^{10} z^{14}

What are the solutions to the equation -16t^2+40t+3=10 ?
The solutions to the equation -16t^2+40t+3=10 are t=(5-3sqrt(2))/4 ,t=(5+3sqrt(2))/4
Find the zeros of -16t^2+40t+3=10
The zeros of -16t^2+40t+3=10 are t=(5-3sqrt(2))/4 ,t=(5+3sqrt(2))/4