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

The solution to 2x^2+2x>= x^2-2x+5 is x = 1

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2x^2+2x>= x^2-2x+5

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

x \leq - 5 or x \geq 1

Solution steps

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

Rewrite in standard form

x^{2} + 4 x - 5 \geq 0

Factor

x^{2} + 4 x - 5 : (x - 1) (x + 5)

(x - 1) (x + 5) \geq 0

Identify the intervals

x \leq - 5 or x \geq 1

24 x^{3} - 9 x^{2} y^{2} - 12

1 1^{\frac{1}{3}} \cdot 12 1^{\frac{1}{3}}

(4)^{\frac{3}{2}}

- 5 + 2

3^{\frac{1}{3}} \cdot 3^{\frac{5}{3}}

What is 2x^2+2x>= x^2-2x+5 ?
The solution to 2x^2+2x>= x^2-2x+5 is x<=-5\lor x>= 1