Solve x^2+12x+35>0

$$x ^ { 2 } + 12 x + 35 > 0$$

Solve for x

$$x\in \left(-\infty,-7\right)\cup \left(-5,\infty\right)$$

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.

$$x^{2}+12x+35=0$$

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 1 for a, 12 for b, and 35 for c in the quadratic formula.

$$x=\frac{-12±\sqrt{12^{2}-4\times 1\times 35}}{2}$$

Do the calculations.

$$x=\frac{-12±2}{2}$$

Solve the equation x=\frac{-12±2}{2} when ± is plus and when ± is minus.

$$x=-5 x=-7$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+5\right)\left(x+7\right)>0$$

For the product to be positive, x+5 and x+7 have to be both negative or both positive. Consider the case when x+5 and x+7 are both negative.

$$x+5<0 x+7<0$$

The solution satisfying both inequalities is x<-7.

$$x<-7$$

Consider the case when x+5 and x+7 are both positive.

$$x+7>0 x+5>0$$

The solution satisfying both inequalities is x>-5.

$$x>-5$$

The final solution is the union of the obtained solutions.

$$x<-7\text{; }x>-5$$