SOLUTION
From the question, the ball's height in meters is represented by the equation
[tex]h=2+25t-5t^2[/tex]Now we are told to find all values of t, for which the ball's height is 12 meters.
This means that h = 12 meters.
So that means
[tex]\begin{gathered} h=2+25t-5t^2=12 \\ So\text{ we have } \\ 12=2+25t-5t^2 \\ 2+25t-5t^2=12 \\ -5t^2+25t+2-12=0 \\ -5t^2+25t-10=0 \\ \text{ multiplying by minus sign we have } \\ -(-5t^2+25t-10=0 \\ 5t^2-25t+10=0 \\ \text{dividing through by 5 we have } \\ t^2-5t+2=0 \end{gathered}[/tex]So now solving the quadratic equation for t, we have
[tex]\begin{gathered} t^2-5t+2=0 \\ t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where a = 1, b = -5, c =2 } \\ t=\frac{-(-5\pm\sqrt[]{(-5)^2-4\times1\times2}}{2\times1} \\ t=\frac{5\pm\sqrt[]{17}}{2} \end{gathered}[/tex]So, either
[tex]\begin{gathered} t=\frac{5+\sqrt[]{17}}{2} \\ t=4.56155281 \\ t=4.56\sec s \end{gathered}[/tex]Or
[tex]\begin{gathered} t=\frac{5-\sqrt[]{17}}{2} \\ t=0.43844718 \\ t=0.44\sec s\text{ } \end{gathered}[/tex]Hence the answer is t = 4.56 secs or 0.44 secs
