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To find the roots of a quadratic equation, plug its coefficients ($$a$$, $$b$$ and $$c$$ ) into the quadratic formula:

$$x=\left(-b\pm sqrt\left(b^2-4ac\right)\right)/\left(2a\right)$$

$$x=\left(-1*-6\pm sqrt\left(-6^2-4*1*9\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-6\pm sqrt\left(36-4*1*9\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-6\pm sqrt\left(36-4*9\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-6\pm sqrt\left(36-36\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-6\pm sqrt\left(0\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-6\pm sqrt\left(0\right)\right)/\left(2\right)$$

$$x=\left(6\pm sqrt\left(0\right)\right)/2$$

to get the result:

$$x=\left(6\pm sqrt\left(0\right)\right)/2$$

Simplify $$0$$ by finding its prime factors:

The prime factorization of $$0$$ is $$0$$

$$x=\left(6\pm 0\right)/2$$

The ± means two roots are possible, but since zero is result of the square root, we have one root:

Separate the equations:
$$x_1=\left(6+0\right)/2$$ and $$x_2=\left(6-0\right)/2$$

$$x_1=\left(6+0\right)/2$$

$$x_1=\left(6+0\right)/2$$

$$x_1=\left(6\right)/2$$

$$x_1=\frac{6}{2}$$

$$x_1=3$$

To find the intervals of a quadratic inequality, we start by finding its parabola.

The roots of the parabola (where it meets the x-axis) are: 3.

Since the $$a$$ coefficient is positive ($$a$$=1), this is a "positive" quadratic inequality and the parabola points upward, like a smile!

If the inequality sign is ≤ or ≥ , then the intervals include the roots and we use a solid line. If the inequality sign is < or > the intervals do not include the roots and we use a dotted line.

Since $$x^2-6x+9<0$$ has a $$<$$ inequality sign, we look for the parabola intervals that are below the x-axis.

Solution: $$$$

Interval notation: $$$$