Solve 3x^2+12x-36<0

$$3 x ^ { 2 } + 12 x - 36 < 0$$

Solve for x

$$x\in \left(-6,2\right)$$

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.

$$3x^{2}+12x-36=0$$

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 3 for a, 12 for b, and -36 for c in the quadratic formula.

$$x=\frac{-12±\sqrt{12^{2}-4\times 3\left(-36\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{-12±24}{6}$$

Solve the equation x=\frac{-12±24}{6} when ± is plus and when ± is minus.

$$x=2 x=-6$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-2\right)\left(x+6\right)<0$$

For the product to be negative, x-2 and x+6 have to be of the opposite signs. Consider the case when x-2 is positive and x+6 is negative.

$$x-2>0 x+6<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+6 is positive and x-2 is negative.

$$x+6>0 x-2<0$$

The solution satisfying both inequalities is x\in \left(-6,2\right).

$$x\in \left(-6,2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-6,2\right)$$