Solve 15z^2+34z+15>0

$$15 z ^ { 2 } + 34 z + 15 > 0$$

Solve for z

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

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.

$$15z^{2}+34z+15=0$$

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 15 for a, 34 for b, and 15 for c in the quadratic formula.

$$z=\frac{-34±\sqrt{34^{2}-4\times 15\times 15}}{2\times 15}$$

Do the calculations.

$$z=\frac{-34±16}{30}$$

Solve the equation z=\frac{-34±16}{30} when ± is plus and when ± is minus.

$$z=-\frac{3}{5} z=-\frac{5}{3}$$

Rewrite the inequality by using the obtained solutions.

$$15\left(z+\frac{3}{5}\right)\left(z+\frac{5}{3}\right)>0$$

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

$$z+\frac{3}{5}<0 z+\frac{5}{3}<0$$

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

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

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

$$z+\frac{5}{3}>0 z+\frac{3}{5}>0$$

The solution satisfying both inequalities is z>-\frac{3}{5}.

$$z>-\frac{3}{5}$$

The final solution is the union of the obtained solutions.

$$z<-\frac{5}{3}\text{; }z>-\frac{3}{5}$$