I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 2.5 : Quadratic Equations - Part I
2. Solve the following quadratic equation by factoring.
\[{x^2} + 15x = - 50\]Show All Steps Hide All Steps
Start SolutionThe first thing we need to do is get everything on one side of the equation and then factor the quadratic.
\[\begin{align*}{x^2} + 15x + 50 & = 0\\ \left( {x + 5} \right)\left( {x + 10} \right) & = 0\end{align*}\] Show Step 2Now all we need to do is use the zero factor property to get,
\[\begin{array}{*{20}{c}}{x + 5 = 0}\\{x = - 5}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{x + 10 = 0}\\{x = - 10}\end{array}\]Therefore the two solutions are : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 5\,\,{\mbox{and }}x = - 10}}\)
We’ll leave it to you to verify that they really are solutions if you’d like to by plugging them back into the equation.