What is 5x^2+9x>= 3x^2+35 ?

The solution to 5x^2+9x>= 3x^2+35 is x = 5/2

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5x^2+9x>= 3x^2+35

5 x^{2} + 9 x \geq 3 x^{2} + 35

x \leq - 7 or x \geq \frac{5}{2}

Interval Notation

(- \infty, - 7] \cup [\frac{5}{2}, \infty)

Decimal Notation

x \leq - 7 or x \geq 2.5

Solution steps

5 x^{2} + 9 x \geq 3 x^{2} + 35

Rewrite in standard form

2 x^{2} + 9 x - 35 \geq 0

Factor

2 x^{2} + 9 x - 35 : (2 x - 5) (x + 7)

(2 x - 5) (x + 7) \geq 0

Identify the intervals

x \leq - 7 or x \geq \frac{5}{2}

1 0^{3}

\frac{1}{\frac{1}{10} + \frac{1}{5}}

4 x^{2} + 32 x + 28

3 - 3 y = 6

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

What is 5x^2+9x>= 3x^2+35 ?
The solution to 5x^2+9x>= 3x^2+35 is x<=-7\lor x>= 5/2