Solve 6x^2-162x+840<0

$$6 { x }^{ 2 } -162x+840 < 0$$

Solve for x

$$x\in \left(7,20\right)$$

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.

$$6x^{2}-162x+840=0$$

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, -162 for b, and 840 for c in the quadratic formula.

$$x=\frac{-\left(-162\right)±\sqrt{\left(-162\right)^{2}-4\times 6\times 840}}{2\times 6}$$

Do the calculations.

$$x=\frac{162±78}{12}$$

Solve the equation x=\frac{162±78}{12} when ± is plus and when ± is minus.

$$x=20 x=7$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-20\right)\left(x-7\right)<0$$

For the product to be negative, x-20 and x-7 have to be of the opposite signs. Consider the case when x-20 is positive and x-7 is negative.

$$x-20>0 x-7<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-7 is positive and x-20 is negative.

$$x-7>0 x-20<0$$

The solution satisfying both inequalities is x\in \left(7,20\right).

$$x\in \left(7,20\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(7,20\right)$$