Solve 6x^2-6>x^2+13x

$$6 x ^ { 2 } - 6 > x ^ { 2 } + 13 x$$

Solve for x

$$x\in \left(-\infty,-\frac{2}{5}\right)\cup \left(3,\infty\right)$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$6x^{2}-6-x^{2}>13x$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-6>13x$$

Subtract 13x from both sides.

$$5x^{2}-6-13x>0$$

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.

$$5x^{2}-6-13x=0$$

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 5 for a, -13 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 5\left(-6\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{13±17}{10}$$

Solve the equation x=\frac{13±17}{10} when ± is plus and when ± is minus.

$$x=3 x=-\frac{2}{5}$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-3\right)\left(x+\frac{2}{5}\right)>0$$

For the product to be positive, x-3 and x+\frac{2}{5} have to be both negative or both positive. Consider the case when x-3 and x+\frac{2}{5} are both negative.

$$x-3<0 x+\frac{2}{5}<0$$

The solution satisfying both inequalities is x<-\frac{2}{5}.

$$x<-\frac{2}{5}$$

Consider the case when x-3 and x+\frac{2}{5} are both positive.

$$x+\frac{2}{5}>0 x-3>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<-\frac{2}{5}\text{; }x>3$$