What is (2x-3)^2+4x^2(x-7)<4(x-2)^3 ?

The solution to (2x-3)^2+4x^2(x-7) 41/60

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(2x-3)^2+4x^2(x-7)<4(x-2)^3

(2 x - 3)^{2} + 4 x^{2} (x - 7) < 4 (x - 2)^{3}

x > \frac{41}{60}

Interval Notation

(\frac{41}{60}, \infty)

Decimal Notation

x > 0.68333 \dots

Solution steps

(2 x - 3)^{2} + 4 x^{2} (x - 7) < 4 (x - 2)^{3}

Rewrite in standard form

- 60 x + 41 < 0

Move

41

to the right side

- 60 x < - 41

Multiply both sides by

- 1

60 x > 41

Divide both sides by

60

x > \frac{41}{60}

- 5 < 20

18 + (70 - 5 x^{2}) = 5.6 x^{2}

x^{2} - x > 6

5 x^{2} + 15 x - 350

2.5 \cdot 1 0^{- 3}

What is (2x-3)^2+4x^2(x-7)<4(x-2)^3 ?
The solution to (2x-3)^2+4x^2(x-7)<4(x-2)^3 is x> 41/60