Solve 12x^2+12x+3>0

$$12 x ^ { 2 } + 12 x + 3 > 0$$

Solve for x

$$x\neq -\frac{1}{2}$$

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.

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

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

$$x=\frac{-12±\sqrt{12^{2}-4\times 12\times 3}}{2\times 12}$$

Do the calculations.

$$x=\frac{-12±0}{24}$$

Solutions are the same.

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

Rewrite the inequality by using the obtained solutions.

$$12\left(x+\frac{1}{2}\right)^{2}>0$$

Inequality holds for x\neq -\frac{1}{2}.

$$x\neq -\frac{1}{2}$$