Solve 6^x^2+6x-1/6>0

$${ 6 }^{ } { x }^{ 2 } +6x- \frac{ 1 }{ 6 } > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{\sqrt{10}}{6}-\frac{1}{2}\right)\cup \left(\frac{\sqrt{10}}{6}-\frac{1}{2},\infty\right)$$

Steps Using the Quadratic Formula

Calculate 6 to the power of 1 and get 6.

$$6x^{2}+6x-\frac{1}{6}>0$$

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.

$$6x^{2}+6x-\frac{1}{6}=0$$

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 6 for a, 6 for b, and -\frac{1}{6} for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 6\left(-\frac{1}{6}\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{-6±2\sqrt{10}}{12}$$

Solve the equation x=\frac{-6±2\sqrt{10}}{12} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{10}}{6}-\frac{1}{2} x=-\frac{\sqrt{10}}{6}-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-\left(\frac{\sqrt{10}}{6}-\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{10}}{6}-\frac{1}{2}\right)\right)>0$$

For the product to be positive, x-\left(\frac{\sqrt{10}}{6}-\frac{1}{2}\right) and x-\left(-\frac{\sqrt{10}}{6}-\frac{1}{2}\right) have to be both negative or both positive. Consider the case when x-\left(\frac{\sqrt{10}}{6}-\frac{1}{2}\right) and x-\left(-\frac{\sqrt{10}}{6}-\frac{1}{2}\right) are both negative.

$$x-\left(\frac{\sqrt{10}}{6}-\frac{1}{2}\right)<0 x-\left(-\frac{\sqrt{10}}{6}-\frac{1}{2}\right)<0$$

The solution satisfying both inequalities is x<-\frac{\sqrt{10}}{6}-\frac{1}{2}.

$$x<-\frac{\sqrt{10}}{6}-\frac{1}{2}$$

Consider the case when x-\left(\frac{\sqrt{10}}{6}-\frac{1}{2}\right) and x-\left(-\frac{\sqrt{10}}{6}-\frac{1}{2}\right) are both positive.

$$x-\left(-\frac{\sqrt{10}}{6}-\frac{1}{2}\right)>0 x-\left(\frac{\sqrt{10}}{6}-\frac{1}{2}\right)>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{10}}{6}-\frac{1}{2}.

$$x>\frac{\sqrt{10}}{6}-\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{\sqrt{10}}{6}-\frac{1}{2}\text{; }x>\frac{\sqrt{10}}{6}-\frac{1}{2}$$