Solve 7x^2-16x+9>0

$$7 { x }^{ 2 } -16x+9 > 0$$

Solve for x

$$x\in \left(-\infty,1\right)\cup \left(\frac{9}{7},\infty\right)$$

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.

$$7x^{2}-16x+9=0$$

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 7 for a, -16 for b, and 9 for c in the quadratic formula.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 7\times 9}}{2\times 7}$$

Do the calculations.

$$x=\frac{16±2}{14}$$

Solve the equation x=\frac{16±2}{14} when ± is plus and when ± is minus.

$$x=\frac{9}{7} x=1$$

Rewrite the inequality by using the obtained solutions.

$$7\left(x-\frac{9}{7}\right)\left(x-1\right)>0$$

For the product to be positive, x-\frac{9}{7} and x-1 have to be both negative or both positive. Consider the case when x-\frac{9}{7} and x-1 are both negative.

$$x-\frac{9}{7}<0 x-1<0$$

The solution satisfying both inequalities is x<1.

$$x<1$$

Consider the case when x-\frac{9}{7} and x-1 are both positive.

$$x-1>0 x-\frac{9}{7}>0$$

The solution satisfying both inequalities is x>\frac{9}{7}.

$$x>\frac{9}{7}$$

The final solution is the union of the obtained solutions.

$$x<1\text{; }x>\frac{9}{7}$$