Explanation
We are to determine the number of real solutions to the quadratic equation
[tex]4u^2-20u+25=0[/tex]To do this we will make use of the quadratic formula:
[tex]\begin{gathered} u_{1,\:2}=\frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:4\cdot \:25}}{2\cdot \:4} \\ \\ \\ u_{1,\:2}=\frac{-\left(-20\right)\pm \sqrt{0}}{2\cdot \:4} \\ \\ u_{1,2}=\frac{-\left(-20\right)\pm0}{2\times\:4} \\ \\ u_1=\frac{20+0}{8}=\frac{20}{8}=\frac{5}{2}=2.5 \\ u_2=\frac{20-0}{8}=\frac{20}{8}=\frac{5}{2}=2.5 \\ \end{gathered}[/tex]Therefore, we can see that it has one real solution which is 2.5
