What is 2^2+5x<-x^2+1 ?

The solution to 2^2+5x<-x^2+1 is (-sqrt(13)-5)/2 <x<(sqrt(13)-5)/2

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2^2+5x<-x^2+1

2^{2} + 5 x < - x^{2} + 1

\frac{- 13 - 5}{2} < x < \frac{13 - 5}{2}

Solution steps

2^{2} + 5 x < - x^{2} + 1

Rewrite in standard form

5 x + 3 + x^{2} < 0

Complete the square

5 x + 3 + x^{2} : (x + \frac{5}{2})^{2} - \frac{13}{4}

(x + \frac{5}{2})^{2} - \frac{13}{4} < 0

Move

\frac{13}{4}

to the right side

(x + \frac{5}{2})^{2} < \frac{13}{4}

For

u^{n} < a

, if

n

is even then

- \frac{13}{4} < x + \frac{5}{2} < \frac{13}{4}

a < u < b

then

a < u and u < b

- \frac{13}{4} < x + \frac{5}{2} and x + \frac{5}{2} < \frac{13}{4}

- \frac{13}{4} < x + \frac{5}{2} : x > \frac{- 13 - 5}{2}

x + \frac{5}{2} < \frac{13}{4} : x < \frac{13 - 5}{2}

Combine the intervals

x > \frac{- 13 - 5}{2} and x < \frac{13 - 5}{2}

Merge Overlapping Intervals

\frac{- 13 - 5}{2} < x < \frac{13 - 5}{2}

3 x^{2} + 12 x - 63 = 0

3 x - 6 \geq - 24

(- 3)^{5}

\frac{1 + 5 i}{5 - i}

- \frac{3}{2} + \frac{5}{2}

What is 2^2+5x<-x^2+1 ?
The solution to 2^2+5x<-x^2+1 is (-sqrt(13)-5)/2 <x<(sqrt(13)-5)/2