Solve -1/4(x^2-4x+3)>0

$$- \frac { 1 } { 4 } ( x ^ { 2 } - 4 x + 3 ) > 0$$

Solve for x

$$x\in \left(1,3\right)$$

Steps Using the Quadratic Formula

Multiply both sides by -4, the reciprocal of -\frac{1}{4}. Since -\frac{1}{4} is negative, the inequality direction is changed. Anything times zero gives zero.

$$x^{2}-4x+3<0$$

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.

$$x^{2}-4x+3=0$$

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, -4 for b, and 3 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{4±2}{2}$$

Solve the equation x=\frac{4±2}{2} when ± is plus and when ± is minus.

$$x=3 x=1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-3\right)\left(x-1\right)<0$$

For the product to be negative, x-3 and x-1 have to be of the opposite signs. Consider the case when x-3 is positive and x-1 is negative.

$$x-3>0 x-1<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-1 is positive and x-3 is negative.

$$x-1>0 x-3<0$$

The solution satisfying both inequalities is x\in \left(1,3\right).

$$x\in \left(1,3\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(1,3\right)$$