Solve 2h^2-7h+6<0

$$2 h ^ { 2 } - 7 h + 6 < 0$$

Solve for h

$$h\in \left(\frac{3}{2},2\right)$$

Steps Using the Quadratic Formula

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.

$$2h^{2}-7h+6=0$$

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 2 for a, -7 for b, and 6 for c in the quadratic formula.

$$h=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Do the calculations.

$$h=\frac{7±1}{4}$$

Solve the equation h=\frac{7±1}{4} when ± is plus and when ± is minus.

$$h=2 h=\frac{3}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(h-2\right)\left(h-\frac{3}{2}\right)<0$$

For the product to be negative, h-2 and h-\frac{3}{2} have to be of the opposite signs. Consider the case when h-2 is positive and h-\frac{3}{2} is negative.

$$h-2>0 h-\frac{3}{2}<0$$

This is false for any h.

$$h\in \emptyset$$

Consider the case when h-\frac{3}{2} is positive and h-2 is negative.

$$h-\frac{3}{2}>0 h-2<0$$

The solution satisfying both inequalities is h\in \left(\frac{3}{2},2\right).

$$h\in \left(\frac{3}{2},2\right)$$

The final solution is the union of the obtained solutions.

$$h\in \left(\frac{3}{2},2\right)$$