What is 5x^2+3x>= 3x^2+20 ?

The solution to 5x^2+3x>= 3x^2+20 is x = 5/2

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5x^2+3x>= 3x^2+20

5 x^{2} + 3 x \geq 3 x^{2} + 20

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

Interval Notation

(- \infty, - 4] \cup [\frac{5}{2}, \infty)

Decimal Notation

x \leq - 4 or x \geq 2.5

Solution steps

5 x^{2} + 3 x \geq 3 x^{2} + 20

Rewrite in standard form

2 x^{2} + 3 x - 20 \geq 0

Factor

2 x^{2} + 3 x - 20 : (2 x - 5) (x + 4)

(2 x - 5) (x + 4) \geq 0

Identify the intervals

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

x^{4} - y^{6}

- 5 t^{2} + 50 t + 5

lo g_{4} (\frac{1}{16}) = x

\frac{x}{2} \leq 6 - x

2 (a + b)^{2} + 7 (a + b) - 4

What is 5x^2+3x>= 3x^2+20 ?
The solution to 5x^2+3x>= 3x^2+20 is x<=-4\lor x>= 5/2