What is (x-3)/4 >(x-2)(x+7)+17 ?

The solution to (x-3)/4 >(x-2)(x+7)+17 is -15/4 <x<-1

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(x-3)/4 >(x-2)(x+7)+17

\frac{x - 3}{4} > (x - 2) (x + 7) + 17

- \frac{15}{4} < x < - 1

Interval Notation

(- \frac{15}{4}, - 1)

Decimal Notation

- 3.75 < x < - 1

Solution steps

\frac{x - 3}{4} > (x - 2) (x + 7) + 17

Rewrite in standard form

- 4 x^{2} - 19 x - 15 > 0

Factor

- 4 x^{2} - 19 x - 15 : - (x + 1) (4 x + 15)

- (x + 1) (4 x + 15) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 1) (4 x + 15)) (- 1) < 0 \cdot (- 1)

Simplify

(x + 1) (4 x + 15) < 0

Identify the intervals

- \frac{15}{4} < x < - 1

- x + 1 = x^{2}

(3^{- 2}) (3^{4})

x^{2} + 2 x + 9 > 0

p^{2} + 27 pq + 50 q^{2}

(\frac{5 ^{2}}{4})^{4}

What is (x-3)/4 >(x-2)(x+7)+17 ?
The solution to (x-3)/4 >(x-2)(x+7)+17 is -15/4 <x<-1