Given the next quadratic equation:
[tex]-x^2+14x+61=0[/tex]we can use the quadratic formula to solve it, as follows:
[tex]\begin{gathered} x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_{1,2}=\frac{-14\pm\sqrt[]{14^2-4\cdot(-1)\cdot61}}{2\cdot(-1)} \\ x_{1,2}=\frac{-14\pm\sqrt[]{196+244}}{-2} \\ x_{1,2}=\frac{-14\pm\sqrt[]{440}}{-2} \\ x_1=\frac{-14+\sqrt[]{440}}{-2}=\frac{-14}{-2}-\frac{\sqrt[]{440}}{2}=7-\sqrt[]{110} \\ x_2=\frac{-14-\sqrt[]{440}}{-2}=\frac{-14}{-2}+\frac{\sqrt[]{440}}{2}=7+\sqrt[]{110} \end{gathered}[/tex]The rounded values (two decimal places) are:
[tex]\begin{gathered} x_1=7-10.49=-3.49 \\ x_2=7+10.49=17.49 \end{gathered}[/tex]Since x is the distance, in ft, from the sprinkler, it cannot be negative, then the answer which makes sense in the context of this problem is 17.49 ft
