Solve a^2-2<a

$$a ^ { 2 } - 2 < a$$

Solve for a

$$a\in \left(-1,2\right)$$

Steps Using the Quadratic Formula

Subtract a from both sides.

$$a^{2}-2-a<0$$

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.

$$a^{2}-2-a=0$$

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 1 for a, -1 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{1±3}{2}$$

Solve the equation a=\frac{1±3}{2} when ± is plus and when ± is minus.

$$a=2 a=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-2\right)\left(a+1\right)<0$$

For the product to be negative, a-2 and a+1 have to be of the opposite signs. Consider the case when a-2 is positive and a+1 is negative.

$$a-2>0 a+1<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+1 is positive and a-2 is negative.

$$a+1>0 a-2<0$$

The solution satisfying both inequalities is a\in \left(-1,2\right).

$$a\in \left(-1,2\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(-1,2\right)$$