What is 6x^2+4x-3>= 0 ?

The solution to 6x^2+4x-3>= 0 is x = (sqrt(11))/(3sqrt(2))-1/3

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6x^2+4x-3>= 0

6 x^{2} + 4 x - 3 \geq 0

x \leq - \frac{11}{3 2} - \frac{1}{3} or x \geq \frac{11}{3 2} - \frac{1}{3}

Interval Notation

(- \infty, - \frac{11}{3 2} - \frac{1}{3}] \cup [\frac{11}{3 2} - \frac{1}{3}, \infty)

Decimal Notation

x \leq - 1.11506 \dots or x \geq 0.44840 \dots

Solution steps

6 x^{2} + 4 x - 3 \geq 0

Complete the square

6 x^{2} + 4 x - 3 : 6 (x + \frac{1}{3})^{2} - \frac{11}{3}

6 (x + \frac{1}{3})^{2} - \frac{11}{3} \geq 0

Move

\frac{11}{3}

to the right side

6 (x + \frac{1}{3})^{2} \geq \frac{11}{3}

Divide both sides by

6

(x + \frac{1}{3})^{2} \geq \frac{11}{18}

For

u^{n} \geq a

, if

n

is even then

x + \frac{1}{3} \leq - \frac{11}{18} or x + \frac{1}{3} \geq \frac{11}{18}

x + \frac{1}{3} \leq - \frac{11}{18} : x \leq - \frac{11}{3 2} - \frac{1}{3}

x + \frac{1}{3} \geq \frac{11}{18} : x \geq \frac{11}{3 2} - \frac{1}{3}

Combine the intervals

x \leq - \frac{11}{3 2} - \frac{1}{3} or x \geq \frac{11}{3 2} - \frac{1}{3}

\frac{1}{4} x - 3 \geq - 1

p^{2} + 3 p - 18

\frac{1}{( \frac{11}{2} )}

λ^{2} + 1 = 0

- \frac{5}{4} + \frac{75}{40} + 100%

What is 6x^2+4x-3>= 0 ?
The solution to 6x^2+4x-3>= 0 is x<=-(sqrt(11))/(3sqrt(2))-1/3 \lor x>= (sqrt(11))/(3sqrt(2))-1/3