Exercise 49 Page 588

Now, we can identify the values of $a,$ $b,$ and $c.$ We see that $a=14,$ $b=\text{-}190,$ and $c=125.$ Let's substitute these values into the Quadratic Formula. The solutions for this equation are $x=\frac{190\pm \sqrt{29100}}{28}.$ Let's separate them into the positive and negative cases.

$x=\frac{190\pm \sqrt{29100}}{28}$
$x_1=\frac{190+\sqrt{29100}}{28}$	$x_2=\frac{190-\sqrt{29100}}{28}$
$x_1=\frac{190}{28}+\frac{\sqrt{29100}}{28}$	$x_2=\frac{190}{28}-\frac{\sqrt{29100}}{28}$
$x_1\approx 12.9$	$x_2\approx 0.7$

Using the Quadratic Formula, we found that the solutions of the given equation are $x_1\approx 12.9$ and $x_2\approx 0.7.$

Note that the other solution $x_1\approx 12.9,$ makes the lengths negative.