What is (x-2)^2>(x+3)(x-3)+8 ?

The solution to (x-2)^2>(x+3)(x-3)+8 is x< 5/4

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(x-2)^2>(x+3)(x-3)+8

(x - 2)^{2} > (x + 3) (x - 3) + 8

x < \frac{5}{4}

Interval Notation

(- \infty, \frac{5}{4})

Decimal Notation

x < 1.25

Solution steps

(x - 2)^{2} > (x + 3) (x - 3) + 8

Rewrite in standard form

- 4 x + 5 > 0

Move

5

to the right side

- 4 x > - 5

Multiply both sides by

- 1

4 x < 5

Divide both sides by

4

x < \frac{5}{4}

- 6.25 x + 262.5 = 250

- 7 < y - 2 < 5

- 2.5 b = 20.5

x^{2} - 10 x + 21 < 0

∣ 3 (x + 1) + 9 ∣ \leq 12