What is x^2+3x-16<0 ?

The solution to x^2+3x-16<0 is (-sqrt(73)-3)/2 <x<(sqrt(73)-3)/2

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x^2+3x-16<0

x^{2} + 3 x - 16 < 0

\frac{- 73 - 3}{2} < x < \frac{73 - 3}{2}

Interval Notation

(\frac{- 73 - 3}{2}, \frac{73 - 3}{2})

Decimal Notation

- 5.77200 \dots < x < 2.77200 \dots

Solution steps

x^{2} + 3 x - 16 < 0

Complete the square

x^{2} + 3 x - 16 : (x + \frac{3}{2})^{2} - \frac{73}{4}

(x + \frac{3}{2})^{2} - \frac{73}{4} < 0

Move

\frac{73}{4}

to the right side

(x + \frac{3}{2})^{2} < \frac{73}{4}

For

u^{n} < a

, if

n

is even then

- \frac{73}{4} < x + \frac{3}{2} < \frac{73}{4}

a < u < b

then

a < u and u < b

- \frac{73}{4} < x + \frac{3}{2} and x + \frac{3}{2} < \frac{73}{4}

- \frac{73}{4} < x + \frac{3}{2} : x > \frac{- 73 - 3}{2}

x + \frac{3}{2} < \frac{73}{4} : x < \frac{73 - 3}{2}

Combine the intervals

x > \frac{- 73 - 3}{2} and x < \frac{73 - 3}{2}

Merge Overlapping Intervals

\frac{- 73 - 3}{2} < x < \frac{73 - 3}{2}

7 ∣ d ∣ \geq 14

(4 + i 4) + (9 - i 9)

(- 16)^{- \frac{3}{2}}

7 \geq \frac{3 - 4 x}{3}

10 x^{3} - 250 x

What is x^2+3x-16<0 ?
The solution to x^2+3x-16<0 is (-sqrt(73)-3)/2 <x<(sqrt(73)-3)/2