A ball is thrown from an initial height of 4 feet with an initial upward velocity of 23 ft/s. The ball's height h (in feet) after 1 seconds is given by the following.h=4+231-167Find all values of 1 for which the ball's height is 12 feet.Round your answer(s) to the nearest hundredth.(If there is more than one answer, use the "or" button.)Please just provide the answer my last tutor lost connection abruptly.

[tex]\begin{gathered} Given\text{ the following equation} \\ h=4+23t-16t^2\text{ } \\ h\text{ = 12 f}eet \\ 12=4+23t-16t^2 \\ \text{Collect the like terms} \\ 12-4=23t-16t^2 \\ 8=23t-16t^2 \\ 23t-16t^2\text{ = 8} \\ -16t^2\text{ + 23t - 8 = 0} \\ \text{ Using the general formula} \\ t\text{ }=\text{ }\frac{-b\pm\sqrt[]{b^2\text{ - 4ac}}}{2a} \\ \text{let a = -16, b = 23, c = -8} \\ t\text{ = }\frac{-23\pm\sqrt[]{(23)^2\text{ - 4}\cdot\text{ }}(-16)\text{ x (-8)}}{2(-16)} \\ t\text{ = }\frac{-23\pm\sqrt[]{529\text{ - 512}}}{-32} \\ t\text{ = }\frac{-23\pm\sqrt[]{17}}{-32} \\ \text{t = -23+}\frac{\sqrt[]{17}}{-32}\text{ or -23-}\frac{\sqrt[]{17}}{-32} \\ t\text{ = -23 }+\text{ 4.12/-32 or t = }\frac{-23\text{ - 4.12}}{-32} \\ t\text{ = }0.59\text{ seconds or t =0.85 seconds} \end{gathered}[/tex]

Therefore, t = 0.59 seconds or t = 0.85 seconds