Solution - Solving quadratic inequalities using the quadratic formula

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

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

$$x=\left(-3\pm sqrt\left(3^2-4*-1*-2\right)\right)/\left(2*-1\right)$$

$$x=\left(-3\pm sqrt\left(9-4*-1*-2\right)\right)/\left(2*-1\right)$$

$$x=\left(-3\pm sqrt\left(9--4*-2\right)\right)/\left(2*-1\right)$$

$$x=\left(-3\pm sqrt\left(9-8\right)\right)/\left(2*-1\right)$$

$$x=\left(-3\pm sqrt\left(1\right)\right)/\left(2*-1\right)$$

$$x=\left(-3\pm sqrt\left(1\right)\right)/\left(-2\right)$$

to get the result:

$$x=\left(-3\pm sqrt\left(1\right)\right)/\left(-2\right)$$

Simplify $$1$$ by finding its prime factors:

The prime factorization of $$1$$ is $$1$$

$$x=\left(-3\pm 1\right)/\left(-2\right)$$

The ± means two roots are possible.

Separate the equations:
$$x_1=\left(-3+1\right)/\left(-2\right)$$ and $$x_2=\left(-3-1\right)/\left(-2\right)$$

$$x_1=\left(-3+1\right)/\left(-2\right)$$

$$x_1=\left(-3+1\right)/\left(-2\right)$$

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

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

$$x_1=1$$

$$x_2=\left(-3-1\right)/\left(-2\right)$$

$$x_2=\left(-3-1\right)/\left(-2\right)$$

$$x_2=\left(-4\right)/\left(-2\right)$$

$$x_2=-\frac{4}{-}2$$

$$x_2=2$$

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: 1, 2.

Since the $$a$$ coefficient is negative ($$a$$=-1), this is a "negative" quadratic inequality and the parabola points downward, like a frown.

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 $$-1x^2+3x-2>0$$ has a $$>$$ inequality sign, we look for the parabola intervals that are above the x-axis.

Solution: $$$$

Interval notation: $$$$