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 2.6 : Quadratic Equations - Part II
13. Use the quadratic formula to solve the following quadratic equation.
\[2{z^2} + z - 72 = {z^2} - 2z + 58\]Show All Steps Hide All Steps
Start SolutionFirst, we need to get the quadratic equation in standard form. This is,
\[{z^2} + 3z - 130 = 0\]In this case we just moved everything from the right side over to the left side.
Show Step 2Now we need to identify the values for the quadratic formula.
\[a = 1\hspace{0.25in}b = 3\hspace{0.25in}c = - 130\] Show Step 3Plugging these into the quadratic formula gives,
\[z = \frac{{ - 3 \pm \sqrt {{{\left( 3 \right)}^2} - 4\left( 1 \right)\left( { - 130} \right)} }}{{2\left( 1 \right)}} = \frac{{ - 3 \pm \sqrt {529} }}{2} = \frac{{ - 3 \pm 23}}{2}\]In this case because there are no roots or complex numbers we can go further to reduce the solutions to “nicer” values.
\[z = \frac{{ - 3 - 23}}{2} = - 13,\,\,\,\,\,z = \frac{{ - 3 + 23}}{2} = 10\]The two solutions to this equation are then : \[\require{bbox} \bbox[2pt,border:1px solid black]{{z = - 13\,\,{\mbox{and }}z = 10}}\] .
As a final comment we can also note that because the solutions were integers we could have also gotten the answer by factoring! The quadratic (once written in standard form) factors as,
\[\left( {z + 13} \right)\left( {z - 10} \right) = 0\]and this clearly would arrive at the same solutions.
The point of all this is to note that if more than one technique can be used it won’t matter which we use. Regardless of the solution technique used you will arrive at the same solutions.