Solve 20x^2+x-1>0

$$20 x ^ { 2 } + x - 1 > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{1}{4}\right)\cup \left(\frac{1}{5},\infty\right)$$

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.

$$20x^{2}+x-1=0$$

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 20 for a, 1 for b, and -1 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 20\left(-1\right)}}{2\times 20}$$

Do the calculations.

$$x=\frac{-1±9}{40}$$

Solve the equation x=\frac{-1±9}{40} when ± is plus and when ± is minus.

$$x=\frac{1}{5} x=-\frac{1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$20\left(x-\frac{1}{5}\right)\left(x+\frac{1}{4}\right)>0$$

For the product to be positive, x-\frac{1}{5} and x+\frac{1}{4} have to be both negative or both positive. Consider the case when x-\frac{1}{5} and x+\frac{1}{4} are both negative.

$$x-\frac{1}{5}<0 x+\frac{1}{4}<0$$

The solution satisfying both inequalities is x<-\frac{1}{4}.

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

Consider the case when x-\frac{1}{5} and x+\frac{1}{4} are both positive.

$$x+\frac{1}{4}>0 x-\frac{1}{5}>0$$

The solution satisfying both inequalities is x>\frac{1}{5}.

$$x>\frac{1}{5}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{4}\text{; }x>\frac{1}{5}$$