Solve 10-3w>8w^2

$$10 - 3 w > 8 w ^ { 2 }$$

Solve for w

$$w\in \left(\frac{-\sqrt{329}-3}{16},\frac{\sqrt{329}-3}{16}\right)$$

Steps Using the Quadratic Formula

Subtract 8w^{2} from both sides.

$$10-3w-8w^{2}>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 10-3w-8w^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-10+3w+8w^{2}<0$$

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.

$$-10+3w+8w^{2}=0$$

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 8 for a, 3 for b, and -10 for c in the quadratic formula.

$$w=\frac{-3±\sqrt{3^{2}-4\times 8\left(-10\right)}}{2\times 8}$$

Do the calculations.

$$w=\frac{-3±\sqrt{329}}{16}$$

Solve the equation w=\frac{-3±\sqrt{329}}{16} when ± is plus and when ± is minus.

$$w=\frac{\sqrt{329}-3}{16} w=\frac{-\sqrt{329}-3}{16}$$

Rewrite the inequality by using the obtained solutions.

$$8\left(w-\frac{\sqrt{329}-3}{16}\right)\left(w-\frac{-\sqrt{329}-3}{16}\right)<0$$

For the product to be negative, w-\frac{\sqrt{329}-3}{16} and w-\frac{-\sqrt{329}-3}{16} have to be of the opposite signs. Consider the case when w-\frac{\sqrt{329}-3}{16} is positive and w-\frac{-\sqrt{329}-3}{16} is negative.

$$w-\frac{\sqrt{329}-3}{16}>0 w-\frac{-\sqrt{329}-3}{16}<0$$

This is false for any w.

$$w\in \emptyset$$

Consider the case when w-\frac{-\sqrt{329}-3}{16} is positive and w-\frac{\sqrt{329}-3}{16} is negative.

$$w-\frac{-\sqrt{329}-3}{16}>0 w-\frac{\sqrt{329}-3}{16}<0$$

The solution satisfying both inequalities is w\in \left(\frac{-\sqrt{329}-3}{16},\frac{\sqrt{329}-3}{16}\right).

$$w\in \left(\frac{-\sqrt{329}-3}{16},\frac{\sqrt{329}-3}{16}\right)$$

The final solution is the union of the obtained solutions.

$$w\in \left(\frac{-\sqrt{329}-3}{16},\frac{\sqrt{329}-3}{16}\right)$$