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:

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

$$x=\left(-13\pm sqrt\left({13}^2-4*56*-3\right)\right)/\left(2*56\right)$$

$$x=\left(-13\pm sqrt\left(169-4*56*-3\right)\right)/\left(2*56\right)$$

$$x=\left(-13\pm sqrt\left(169-224*-3\right)\right)/\left(2*56\right)$$

$$x=\left(-13\pm sqrt\left(169--672\right)\right)/\left(2*56\right)$$

$$x=\left(-13\pm sqrt\left(169+672\right)\right)/\left(2*56\right)$$

$$x=\left(-13\pm sqrt\left(841\right)\right)/\left(2*56\right)$$

$$x=\left(-13\pm sqrt\left(841\right)\right)/\left(112\right)$$

to get the result:

$$x=\left(-13\pm sqrt\left(841\right)\right)/112$$

Simplify $$841$$ by finding its prime factors:

The prime factorization of $$841$$ is $${29}^2$$

$$x=\left(-13\pm 29\right)/112$$

The ± means two roots are possible.

Separate the equations:
$$x_1=\left(-13+29\right)/112$$ and $$x_2=\left(-13-29\right)/112$$

$$x_1=\left(-13+29\right)/112$$

$$x_1=\left(-13+29\right)/112$$

$$x_1=\left(16\right)/112$$

$$x_1=\frac{16}{112}$$

$$x_1=0.143$$

$$x_2=\left(-13-29\right)/112$$

$$x_2=\left(-13-29\right)/112$$

$$x_2=\left(-42\right)/112$$

$$x_2=-\frac{42}{112}$$

$$x_2=-0.375$$

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: -0.375, 0.143.

Since the $$a$$ coefficient is positive ($$a$$=56), 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 $$56x^2+13x-3<0$$ has a $$<$$ inequality sign, we look for the parabola intervals that are below the x-axis.

Solution: $$$$

Interval notation: $$$$