Solve 6x^2-5<x

$$6 x ^ { 2 } - 5 < x$$

Solve for x

$$x\in \left(-\frac{5}{6},1\right)$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$6x^{2}-5-x<0$$

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.

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

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, -1 for b, and -5 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{1±11}{12}$$

Solve the equation x=\frac{1±11}{12} when ± is plus and when ± is minus.

$$x=1 x=-\frac{5}{6}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-1\right)\left(x+\frac{5}{6}\right)<0$$

For the product to be negative, x-1 and x+\frac{5}{6} have to be of the opposite signs. Consider the case when x-1 is positive and x+\frac{5}{6} is negative.

$$x-1>0 x+\frac{5}{6}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{5}{6} is positive and x-1 is negative.

$$x+\frac{5}{6}>0 x-1<0$$

The solution satisfying both inequalities is x\in \left(-\frac{5}{6},1\right).

$$x\in \left(-\frac{5}{6},1\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{5}{6},1\right)$$