Question

Solve using the quadratic formula

-4.9t^2 + 20t + 10

Answer 1

`\bf ~~~~~~~~~~~~\textit{quadratic formula} \\\\ \begin{array}{lcccl} y=& -4.9 t^2& +20t& +10\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t=\cfrac{-20\pm√(20^2-4(-4.9)(10))}{2(-4.9)}\implies t=\cfrac{-20\pm√(400+196)}{-9.8} \\\\\\ t=\cfrac{-20\pm√(596)}{-9.8}\implies t=\cfrac{20\mp√(596)}{9.8}\implies t\approx \begin{cases} -0.4503174725\\ 4.53195012566 \end{cases}`