What is x^2+5x-7>= 0 ?

The solution to x^2+5x-7>= 0 is x = (sqrt(53)-5)/2

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x^2+5x-7>= 0

x^{2} + 5 x - 7 \geq 0

x \leq \frac{- 53 - 5}{2} or x \geq \frac{53 - 5}{2}

Interval Notation

(- \infty, \frac{- 53 - 5}{2}] \cup [\frac{53 - 5}{2}, \infty)

Decimal Notation

x \leq - 6.14005 \dots or x \geq 1.14005 \dots

Solution steps

x^{2} + 5 x - 7 \geq 0

Complete the square

x^{2} + 5 x - 7 : (x + \frac{5}{2})^{2} - \frac{53}{4}

(x + \frac{5}{2})^{2} - \frac{53}{4} \geq 0

Move

\frac{53}{4}

to the right side

(x + \frac{5}{2})^{2} \geq \frac{53}{4}

For

u^{n} \geq a

, if

n

is even then

x + \frac{5}{2} \leq - \frac{53}{4} or x + \frac{5}{2} \geq \frac{53}{4}

x + \frac{5}{2} \leq - \frac{53}{4} : x \leq \frac{- 53 - 5}{2}

x + \frac{5}{2} \geq \frac{53}{4} : x \geq \frac{53 - 5}{2}

Combine the intervals

x \leq \frac{- 53 - 5}{2} or x \geq \frac{53 - 5}{2}

- \frac{12}{7} = - 6 y

\frac{π}{2} + \frac{3 π}{2}

x^{2} + x^{2} = 36

\frac{3 ^{3}}{3 !}

3 (n + 1) = 15

What is x^2+5x-7>= 0 ?
The solution to x^2+5x-7>= 0 is x<= (-sqrt(53)-5)/2 \lor x>= (sqrt(53)-5)/2