Here we consider inequalities of the form ax^2+bx+c>0 or ax^2+bx+c<0, where a!=0.

Fact. If discriminant D=b^2-4ac of quadratic function ax^2+bx+c is negative and a>0, then for all x ax^2+bx+c>0.

Now, let's consider case when D>=0. To solve inequality ax^2+bx+c>0 (or ax^2+bx+c<0) we need to factor ax^2+bx+c: ax^2+bx+c=a(x-x_1)(x-x_2) and then divide both sides of inequality a(x-x_1)(x-x_2)>0 (or a(x-x_1)(x-x_2)<0) by a, changing sign of inequality if a<0, i.e. obtaining inequality (x-x_1)(x-x_2)>0 (or (x-x_1)(x-x_2)<0). Now we need only to use fact that product of two numbers is positive (negative), if multipliers have same (different) signs.

Example 1. Solve 2x^2+5x+2>0.

Let's find roots of 2x^2+5x+2.

From equation 2x^2+5x+2=0 we found that x_1=-2, x_2=-1/2.

Thus, 2x^2+5x+2=2(x+2)(x+1/2) and we obtain inequality 2(x+2)(x+1/2)>0, i.e. (x+2)(x+1/2)>0.

This means that x+2 and x+1/2 should have same signs, i.e.

{(x+2 >0),(x+1/2>0):} or {(x+2<0),(x+1/2<0):}.

From first system we have that x> -1/2, from second system x< -2. Taking union of these intervals we have that solution of the initial quadratic inequality is (-oo,-2)uu(-0.5;+oo) (uu is union sign).

Example 2. Solve 5-3x>=2x^2.

First we need to transform given inequality into standard form: 2x^2+3x-5<=0.

Let's find roots of 2x^2+3x-5.

From equation 2x^2+3x-5=0 we found that x_1=-5/2, x_2=1.

Thus, 2x^2+3x-5=2(x+5/2)(x-1) and we obtain inequality 2(x+5/2)(x-1)<=0, i.e. (x+5/2)(x-1)<=0.

This means that x+5/2 and x-1 should have different signs, i.e.

{(x+5/2 <=0),(x-1>=0):} or {(x+5/2>=0),(x-1<=0):}.

First system doesn't have solutions, from second system -5/2<=x<=1. Therefore solution of the initial quadratic inequality is -5/2<=x<=1.