Solve 10x^2+11x+3>0

$$10 { x }^{ 2 } +11x+3 > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{3}{5}\right)\cup \left(-\frac{1}{2},\infty\right)$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$10x^{2}+11x+3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 10 for a, 11 for b, and 3 for c in the quadratic formula.

$$x=\frac{-11±\sqrt{11^{2}-4\times 10\times 3}}{2\times 10}$$

Do the calculations.

$$x=\frac{-11±1}{20}$$

Solve the equation x=\frac{-11±1}{20} when ± is plus and when ± is minus.

$$x=-\frac{1}{2} x=-\frac{3}{5}$$

Rewrite the inequality by using the obtained solutions.

$$10\left(x+\frac{1}{2}\right)\left(x+\frac{3}{5}\right)>0$$

For the product to be positive, x+\frac{1}{2} and x+\frac{3}{5} have to be both negative or both positive. Consider the case when x+\frac{1}{2} and x+\frac{3}{5} are both negative.

$$x+\frac{1}{2}<0 x+\frac{3}{5}<0$$

The solution satisfying both inequalities is x<-\frac{3}{5}.

$$x<-\frac{3}{5}$$

Consider the case when x+\frac{1}{2} and x+\frac{3}{5} are both positive.

$$x+\frac{3}{5}>0 x+\frac{1}{2}>0$$

The solution satisfying both inequalities is x>-\frac{1}{2}.

$$x>-\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{3}{5}\text{; }x>-\frac{1}{2}$$