Solve -5x^2+7x-2<0

$$- 5 x ^ { 2 } + 7 x - 2 < 0$$

Solve for x

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

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -5x^{2}+7x-2 positive. Since -1 is negative, the inequality direction is changed.

$$5x^{2}-7x+2>0$$

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.

$$5x^{2}-7x+2=0$$

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 5 for a, -7 for b, and 2 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 5\times 2}}{2\times 5}$$

Do the calculations.

$$x=\frac{7±3}{10}$$

Solve the equation x=\frac{7±3}{10} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$5\left(x-1\right)\left(x-\frac{2}{5}\right)>0$$

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

$$x-1<0 x-\frac{2}{5}<0$$

The solution satisfying both inequalities is x<\frac{2}{5}.

$$x<\frac{2}{5}$$

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

$$x-\frac{2}{5}>0 x-1>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

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