What is 2222-95(t)+1(t)^2<0 ?

The solution to 2222-95(t)+1(t)^2<0 is (-sqrt(137)+95)/2 <t<(sqrt(137)+95)/2

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2222-95(t)+1(t)^2<0

2222 - 95 (t) + 1 (t)^{2} < 0

\frac{- 137 + 95}{2} < t < \frac{137 + 95}{2}

Solution steps

2222 - 95 t + 1 \cdot t^{2} < 0

Rewrite in standard form

2222 - 95 t + t^{2} < 0

Complete the square

2222 - 95 t + t^{2} : (t - \frac{95}{2})^{2} - \frac{137}{4}

(t - \frac{95}{2})^{2} - \frac{137}{4} < 0

Move

\frac{137}{4}

to the right side

(t - \frac{95}{2})^{2} < \frac{137}{4}

For

u^{n} < a

, if

n

is even then

- \frac{137}{4} < t - \frac{95}{2} < \frac{137}{4}

a < u < b

then

a < u and u < b

- \frac{137}{4} < t - \frac{95}{2} and t - \frac{95}{2} < \frac{137}{4}

- \frac{137}{4} < t - \frac{95}{2} : t > \frac{- 137 + 95}{2}

t - \frac{95}{2} < \frac{137}{4} : t < \frac{137 + 95}{2}

Combine the intervals

t > \frac{- 137 + 95}{2} and t < \frac{137 + 95}{2}

Merge Overlapping Intervals

\frac{- 137 + 95}{2} < t < \frac{137 + 95}{2}

x^{3} - a^{3}

81 - 49 x^{2}

4 w^{2} + 12 w = - 5

5 - 2 x < 7

13 - \frac{3}{2} x = 37

What is 2222-95(t)+1(t)^2<0 ?
The solution to 2222-95(t)+1(t)^2<0 is (-sqrt(137)+95)/2 <t<(sqrt(137)+95)/2