What is x^2+3x-5>= 0 ?

The solution to x^2+3x-5>= 0 is x = (sqrt(29)-3)/2

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x^2+3x-5>= 0

x^{2} + 3 x - 5 \geq 0

x \leq \frac{- 29 - 3}{2} or x \geq \frac{29 - 3}{2}

Interval Notation

(- \infty, \frac{- 29 - 3}{2}] \cup [\frac{29 - 3}{2}, \infty)

Decimal Notation

x \leq - 4.19258 \dots or x \geq 1.19258 \dots

Solution steps

x^{2} + 3 x - 5 \geq 0

Complete the square

x^{2} + 3 x - 5 : (x + \frac{3}{2})^{2} - \frac{29}{4}

(x + \frac{3}{2})^{2} - \frac{29}{4} \geq 0

Move

\frac{29}{4}

to the right side

(x + \frac{3}{2})^{2} \geq \frac{29}{4}

For

u^{n} \geq a

, if

n

is even then

x + \frac{3}{2} \leq - \frac{29}{4} or x + \frac{3}{2} \geq \frac{29}{4}

x + \frac{3}{2} \leq - \frac{29}{4} : x \leq \frac{- 29 - 3}{2}

x + \frac{3}{2} \geq \frac{29}{4} : x \geq \frac{29 - 3}{2}

Combine the intervals

x \leq \frac{- 29 - 3}{2} or x \geq \frac{29 - 3}{2}

\frac{k}{- 11} = - 3

4^{- 1} = \frac{1}{4}

(\frac{4 x ^{7}}{6 x})^{2}

x^{2} + 9 x y - 36 y^{2}

12 x - 15 = 6 x + 9

What is x^2+3x-5>= 0 ?
The solution to x^2+3x-5>= 0 is x<= (-sqrt(29)-3)/2 \lor x>= (sqrt(29)-3)/2