Solve for t. d=−16t2+4t t=12±41−4d‾‾‾‾‾‾√ t=8±1−4d‾‾‾‾‾‾√ t=18±1−4d√8 t=12±41−4d√2
Accepted Solution
A:
Answer:[tex]t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)[/tex]Step-by-step explanation:Rearranging your quadratic to standard form, you get ... 16t^2 -4t +d = 0For the quadratic formula, you have ... a = 16 b = -4 c = dso the solution is ...[tex]t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(16)(d)}}{2(16)}=\dfrac{4\pm 4\sqrt{1-4d}}{32}\\\\t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)[/tex]